Greatest common substring












24












$begingroup$


Create a program or function which takes a list of strings as input, and outputs the longest string that is a substring of all input strings. If there are several substrings of equal length, and no longer substring, output any one of them.




  • This may mean outputting the empty string.

  • If there are several valid outputs, you may output any one of them. You are not required to give consistent output for a given input so long as the output is always valid.

  • There will always be at least one string in the input, but there might not be a non-empty string.

  • All printable ASCII characters may appear in the input. You may assume those are the only characters that appear.

  • You may take input or produce output by any of the default methods.


  • Standard loopholes aren't allowed.

  • This is code-golf - the fewer bytes of code, the better.


Test cases:



[Inputs] -> [Valid outputs (choose one)]



["hello", "'ello"] -> ["ello"]
["very", "much", "different"] -> [""]
["empty", "", "STRING"] -> [""]
["identical", "identical"] -> ["identical"]
["string", "stRIng"] -> ["st", "ng"]
["this one", "is a substring of this one"] -> ["this one"]
["just one"] -> ["just one"]
["", "", ""] -> [""]
["many outputs", "stuptuo ynam"] -> ["m", "a", "n", "y", " ", "o", "u", "t", "p", "s"]
["many inputs", "any inputs", "ny iii", "yanny"] -> ["ny"]
["%%not&", "ju&#st", "[&]alpha_numeric"] -> ["&"]









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$endgroup$












  • $begingroup$
    Possible duplicate
    $endgroup$
    – Adám
    yesterday






  • 1




    $begingroup$
    @Adám That question asks for the longest common subsequence, not substring.
    $endgroup$
    – Doorknob
    yesterday






  • 1




    $begingroup$
    Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
    $endgroup$
    – Embodiment of Ignorance
    yesterday










  • $begingroup$
    @EmbodimentofIgnorance All printable ASCII characters can appear in the input.
    $endgroup$
    – Sara J
    yesterday






  • 1




    $begingroup$
    @Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
    $endgroup$
    – Sara J
    20 hours ago
















24












$begingroup$


Create a program or function which takes a list of strings as input, and outputs the longest string that is a substring of all input strings. If there are several substrings of equal length, and no longer substring, output any one of them.




  • This may mean outputting the empty string.

  • If there are several valid outputs, you may output any one of them. You are not required to give consistent output for a given input so long as the output is always valid.

  • There will always be at least one string in the input, but there might not be a non-empty string.

  • All printable ASCII characters may appear in the input. You may assume those are the only characters that appear.

  • You may take input or produce output by any of the default methods.


  • Standard loopholes aren't allowed.

  • This is code-golf - the fewer bytes of code, the better.


Test cases:



[Inputs] -> [Valid outputs (choose one)]



["hello", "'ello"] -> ["ello"]
["very", "much", "different"] -> [""]
["empty", "", "STRING"] -> [""]
["identical", "identical"] -> ["identical"]
["string", "stRIng"] -> ["st", "ng"]
["this one", "is a substring of this one"] -> ["this one"]
["just one"] -> ["just one"]
["", "", ""] -> [""]
["many outputs", "stuptuo ynam"] -> ["m", "a", "n", "y", " ", "o", "u", "t", "p", "s"]
["many inputs", "any inputs", "ny iii", "yanny"] -> ["ny"]
["%%not&", "ju&#st", "[&]alpha_numeric"] -> ["&"]









share|improve this question











$endgroup$












  • $begingroup$
    Possible duplicate
    $endgroup$
    – Adám
    yesterday






  • 1




    $begingroup$
    @Adám That question asks for the longest common subsequence, not substring.
    $endgroup$
    – Doorknob
    yesterday






  • 1




    $begingroup$
    Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
    $endgroup$
    – Embodiment of Ignorance
    yesterday










  • $begingroup$
    @EmbodimentofIgnorance All printable ASCII characters can appear in the input.
    $endgroup$
    – Sara J
    yesterday






  • 1




    $begingroup$
    @Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
    $endgroup$
    – Sara J
    20 hours ago














24












24








24


1



$begingroup$


Create a program or function which takes a list of strings as input, and outputs the longest string that is a substring of all input strings. If there are several substrings of equal length, and no longer substring, output any one of them.




  • This may mean outputting the empty string.

  • If there are several valid outputs, you may output any one of them. You are not required to give consistent output for a given input so long as the output is always valid.

  • There will always be at least one string in the input, but there might not be a non-empty string.

  • All printable ASCII characters may appear in the input. You may assume those are the only characters that appear.

  • You may take input or produce output by any of the default methods.


  • Standard loopholes aren't allowed.

  • This is code-golf - the fewer bytes of code, the better.


Test cases:



[Inputs] -> [Valid outputs (choose one)]



["hello", "'ello"] -> ["ello"]
["very", "much", "different"] -> [""]
["empty", "", "STRING"] -> [""]
["identical", "identical"] -> ["identical"]
["string", "stRIng"] -> ["st", "ng"]
["this one", "is a substring of this one"] -> ["this one"]
["just one"] -> ["just one"]
["", "", ""] -> [""]
["many outputs", "stuptuo ynam"] -> ["m", "a", "n", "y", " ", "o", "u", "t", "p", "s"]
["many inputs", "any inputs", "ny iii", "yanny"] -> ["ny"]
["%%not&", "ju&#st", "[&]alpha_numeric"] -> ["&"]









share|improve this question











$endgroup$




Create a program or function which takes a list of strings as input, and outputs the longest string that is a substring of all input strings. If there are several substrings of equal length, and no longer substring, output any one of them.




  • This may mean outputting the empty string.

  • If there are several valid outputs, you may output any one of them. You are not required to give consistent output for a given input so long as the output is always valid.

  • There will always be at least one string in the input, but there might not be a non-empty string.

  • All printable ASCII characters may appear in the input. You may assume those are the only characters that appear.

  • You may take input or produce output by any of the default methods.


  • Standard loopholes aren't allowed.

  • This is code-golf - the fewer bytes of code, the better.


Test cases:



[Inputs] -> [Valid outputs (choose one)]



["hello", "'ello"] -> ["ello"]
["very", "much", "different"] -> [""]
["empty", "", "STRING"] -> [""]
["identical", "identical"] -> ["identical"]
["string", "stRIng"] -> ["st", "ng"]
["this one", "is a substring of this one"] -> ["this one"]
["just one"] -> ["just one"]
["", "", ""] -> [""]
["many outputs", "stuptuo ynam"] -> ["m", "a", "n", "y", " ", "o", "u", "t", "p", "s"]
["many inputs", "any inputs", "ny iii", "yanny"] -> ["ny"]
["%%not&", "ju&#st", "[&]alpha_numeric"] -> ["&"]






code-golf string subsequence






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share|improve this question








edited 14 hours ago









Toby Speight

4,47711535




4,47711535










asked yesterday









Sara JSara J

2818




2818












  • $begingroup$
    Possible duplicate
    $endgroup$
    – Adám
    yesterday






  • 1




    $begingroup$
    @Adám That question asks for the longest common subsequence, not substring.
    $endgroup$
    – Doorknob
    yesterday






  • 1




    $begingroup$
    Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
    $endgroup$
    – Embodiment of Ignorance
    yesterday










  • $begingroup$
    @EmbodimentofIgnorance All printable ASCII characters can appear in the input.
    $endgroup$
    – Sara J
    yesterday






  • 1




    $begingroup$
    @Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
    $endgroup$
    – Sara J
    20 hours ago


















  • $begingroup$
    Possible duplicate
    $endgroup$
    – Adám
    yesterday






  • 1




    $begingroup$
    @Adám That question asks for the longest common subsequence, not substring.
    $endgroup$
    – Doorknob
    yesterday






  • 1




    $begingroup$
    Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
    $endgroup$
    – Embodiment of Ignorance
    yesterday










  • $begingroup$
    @EmbodimentofIgnorance All printable ASCII characters can appear in the input.
    $endgroup$
    – Sara J
    yesterday






  • 1




    $begingroup$
    @Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
    $endgroup$
    – Sara J
    20 hours ago
















$begingroup$
Possible duplicate
$endgroup$
– Adám
yesterday




$begingroup$
Possible duplicate
$endgroup$
– Adám
yesterday




1




1




$begingroup$
@Adám That question asks for the longest common subsequence, not substring.
$endgroup$
– Doorknob
yesterday




$begingroup$
@Adám That question asks for the longest common subsequence, not substring.
$endgroup$
– Doorknob
yesterday




1




1




$begingroup$
Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
$endgroup$
– Embodiment of Ignorance
yesterday




$begingroup$
Will the strings be only alphanumeric, or alphabetic, or only printable-ascii?
$endgroup$
– Embodiment of Ignorance
yesterday












$begingroup$
@EmbodimentofIgnorance All printable ASCII characters can appear in the input.
$endgroup$
– Sara J
yesterday




$begingroup$
@EmbodimentofIgnorance All printable ASCII characters can appear in the input.
$endgroup$
– Sara J
yesterday




1




1




$begingroup$
@Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
$endgroup$
– Sara J
20 hours ago




$begingroup$
@Shaggy Generally, no. If the two can be distinguished, undefined implies there's no valid output string. If the empty string (or any other string) is a valid output, claiming there is no valid output is incorrect.
$endgroup$
– Sara J
20 hours ago










23 Answers
23






active

oldest

votes


















8












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Brachylog (v2), 3 9 bytes



{sᵛ}ᶠlᵒtw


Try it online!



Full program. Input from standard input (as a JSON-style list of strings), output to standard output.



Explanation



{sᵛ}ᶠlᵒtw
s Find a substring
ᵛ of every element {of the input}; the same one for each
{ }ᶠ Convert generator to list
lᵒt Take list element with maximum length
w Output it


Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)



Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.






share|improve this answer











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  • 1




    $begingroup$
    I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
    $endgroup$
    – t-clausen.dk
    18 hours ago










  • $begingroup$
    Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
    $endgroup$
    – ais523
    18 hours ago












  • $begingroup$
    @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
    $endgroup$
    – Kroppeb
    13 hours ago



















7












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Python 2, 82 bytes





f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)


Try it online!



Takes input splatted. Will time out for inputs where the first string is long.



The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.





Python 2, 94 bytes





lambda l:max(set.intersection(*map(g,l)),key=len)
g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}


Try it online!



A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.






share|improve this answer









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    4












    $begingroup$


    Ruby 2.6, 76 59 54 bytes





    ->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}


    Try it online! - Ruby 2.5 version (56 bytes)



    How?



    Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.



    Thanks Kirill L for -2 bytes and histocrat for another -2






    share|improve this answer











    $endgroup$









    • 1




      $begingroup$
      You might as well say that it is Ruby 2.6 and get rid of -1 :)
      $endgroup$
      – Kirill L.
      14 hours ago



















    4












    $begingroup$


    Zsh, 126 123 bytes



    3 bytes saved via arithmetic for



    for l in $@
    {a= i=
    for ((;i++<$#l**2;)){a+=($l[1+i/$#l,1+i%$#l])}
    b=(${${b-$a}:*a})}
    for s in $b;{(($#x<$#s))&&x=$s;}
    <<<$x


    Try it online!



    We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.



    for l in $@; {
    a= # empty a
    for (( ; i++ < $#l**2; )) { # compound double loop using div/mod
    a+=( $l[1+i/$#l,1+i%$#l] ) # append to a all possible substrings of the given line
    }
    b=( ${${b-$a}:*a} )
    # ${b-$a} # if b is unset substitute $a
    # ${ :*a} # take common elements of ${b-$a} and $a
    # b=( ) # set b to those elements
    }
    for s in $b; { # for every common substring
    (( $#x < $#s )) && x=$s # if the current word is longer, use it
    }
    <<<$x # print to stdout





    share|improve this answer










    New contributor




    GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$





















      3












      $begingroup$


      Perl 6, 62 bytes





      {~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}


      Try it online!



      I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.






      share|improve this answer









      $endgroup$









      • 3




        $begingroup$
        max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
        $endgroup$
        – Ven
        21 hours ago





















      3












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      05AB1E, 14 9 bytes



      €Œ.«Ãõªéθ


      -5 bytes thanks to @Adnan.



      Try it online or verify all test cases.



      Explanation:





      €Œ         # Get the substring of each string in the (implicit) input-list
      .« # Right-reduce this list of list of strings by:
      Ã # Only keep all the strings that are present in both list of strings
      õª # Append an empty string to the remaining list of strings
      é # Sort by length
      θ # And pop and push its last item (which is output implicitly as result)





      share|improve this answer











      $endgroup$









      • 1




        $begingroup$
        I think €Œ.«Ãõªéθ should work for 9 bytes.
        $endgroup$
        – Adnan
        16 hours ago










      • $begingroup$
        @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
        $endgroup$
        – Kevin Cruijssen
        16 hours ago










      • $begingroup$
        Actually, I think €Œ.«ÃéθJ should work for 8.
        $endgroup$
        – Adnan
        10 hours ago



















      3












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      Jelly, 12 6 bytes



      Ẇ€f/ṫ0


      Try it online!



      Thanks to @JonathanAllan for saving 6 bytes!






      share|improve this answer











      $endgroup$













      • $begingroup$
        I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
        $endgroup$
        – Jonathan Allan
        18 hours ago










      • $begingroup$
        also I think œ&/ may be replaced by f/ here saving another
        $endgroup$
        – Jonathan Allan
        18 hours ago










      • $begingroup$
        I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
        $endgroup$
        – Jonathan Allan
        14 hours ago












      • $begingroup$
        @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
        $endgroup$
        – Nick Kennedy
        14 hours ago










      • $begingroup$
        Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
        $endgroup$
        – Jonathan Allan
        14 hours ago



















      3












      $begingroup$


      R, 119 116 108 106 bytes





      function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]


      Try it online!



      Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.



      -3 bytes thanks to Kirill L.



      -8 bytes using lapply instead of Map



      -2 bytes thanks to Kirill L. again, removing braces






      share|improve this answer











      $endgroup$













      • $begingroup$
        I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
        $endgroup$
        – Kirill L.
        14 hours ago










      • $begingroup$
        @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
        $endgroup$
        – Giuseppe
        14 hours ago












      • $begingroup$
        You can also drop braces for another -2
        $endgroup$
        – Kirill L.
        8 hours ago










      • $begingroup$
        @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
        $endgroup$
        – Giuseppe
        8 hours ago










      • $begingroup$
        that's the problem with golfing a language you use every day
        $endgroup$
        – MickyT
        8 hours ago



















      2












      $begingroup$


      Haskell, 80 bytes





      import Data.List
      f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]


      Try it online!



      Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.






      share|improve this answer











      $endgroup$





















        2












        $begingroup$

        C#, 320 bytes



        //method that returns all possible substrings from a string
        IEnumerable<string>f(string s){for(int i=0;i<s.Length;i++)for(int j=i;j<s.Length;j++)yield return s.Substring(i,j-i+1);}

        //main method
        void g(List<string>l){Console.WriteLine(string.Join(",",l.Select(s=>f(s)).Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x=>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));}


        Try it online!



        Couldn't make it work for https://tio.run/#cs-csc, would be really grateful if someone showed me how to input lists on that site!



        The result can be outputted as a collection rather than a string, which would save a few bytes, but it seems like in C#, unlike with other languages, it won't be readable through Console.Writeline, would that still be a valid answer or is there a different workaround?






        share|improve this answer








        New contributor




        Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$













        • $begingroup$
          You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
          $endgroup$
          – Expired Data
          14 hours ago












        • $begingroup$
          Here's the solution golfed down some bytes 215 bytes
          $endgroup$
          – Expired Data
          13 hours ago



















        2












        $begingroup$


        Japt v2.0a0 -hF, 8 bytes



        Îã f@eøX


        Thanks to Shaggy for saving 3 bytes



        Try it



        Îã              //Generate all substrings of the first string
        f@ //Filter; keep the substrings that satisfy the following predicate:
        e // If all strings of the input...
        øX // Contain this substring, then keep it
        -h //Take last element
        -F //If last element is undefined, default to empty string





        share|improve this answer











        $endgroup$













        • $begingroup$
          You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
          $endgroup$
          – Shaggy
          19 hours ago



















        1












        $begingroup$


        Python 3, 137 bytes





        def a(b):c=[[d[f:e]for e in range(len(d)+1)for f in range(e+1)]for d in b];return max([i for i in c[0]if all(i in j for j in c)],key=len)


        Try it online!






        share|improve this answer










        New contributor




        Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        $endgroup$













        • $begingroup$
          You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
          $endgroup$
          – Shieru Asakoto
          yesterday










        • $begingroup$
          @JoKing tio.run/…
          $endgroup$
          – Artemis Fowl
          yesterday










        • $begingroup$
          Right, here's a fixed version of the 135 byte program
          $endgroup$
          – Jo King
          yesterday





















        1












        $begingroup$


        Python 2, 103 bytes





        lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)


        Try it online!



        This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.






        share|improve this answer











        $endgroup$













        • $begingroup$
          1 byte shorter with set.intersection.
          $endgroup$
          – ovs
          19 hours ago



















        1












        $begingroup$


        Japt -h, 8 bytes



        (I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)



        mã rf iP


        Try it or run all test cases



        mã rf iP     :Implicit input of array
        m :Map
        ã : Substrings
        r :Reduce by
        f : Filter, keeping only elements that appear in both arrays
        i :Prepend
        P : An empty string
        :Implicit output of last element





        share|improve this answer











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          1












          $begingroup$

          TSQL query, 154 bytes



          USE master
          DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)
          INSERT @ values('string'),('stRIng');

          SELECT top 1x FROM(SELECT
          distinct substring(a,f.number,g.number)x,i
          FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D
          GROUP BY x ORDER BY-sum(i),-len(x)


          Try it online






          share|improve this answer











          $endgroup$





















            1












            $begingroup$

            Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes



            may be golfed more certainly



            map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h


            TIO






            share|improve this answer











            $endgroup$





















              0












              $begingroup$


              Java (JDK), 176 bytes





              a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}


              Try it online!



              This is a rather naive implementation.






              share|improve this answer









              $endgroup$





















                0












                $begingroup$


                C# (Visual C# Interactive Compiler), 147 145 bytes





                a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}


                Try it online!






                share|improve this answer











                $endgroup$





















                  0












                  $begingroup$

                  JavaScript (ES6),  98  92 bytes





                  a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])


                  Try it online!






                  share|improve this answer











                  $endgroup$





















                    0












                    $begingroup$


                    Red, 266 174 bytes



                    func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u
                    next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]


                    Try it online!



                    Changed the recursion to iteration and got rid of the sorting.






                    share|improve this answer











                    $endgroup$





















                      0












                      $begingroup$


                      Perl 5, 87 bytes





                      my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r


                      Try it online!






                      share|improve this answer









                      $endgroup$





















                        0












                        $begingroup$


                        Retina 0.8.2, 48 bytes



                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)
                        O#$^`
                        $.&
                        1G`


                        Try it online! Explanation:



                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)


                        For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.



                        O#$^`
                        $.&


                        Sort the substrings in reverse order of length.



                        1G`


                        Keep only the first, i.e. the longest substring.






                        share|improve this answer











                        $endgroup$





















                          0












                          $begingroup$


                          JavaScript (Node.js), 106 bytes





                          a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.indexOf(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)


                          Try it online!





                          a=>(                      // Main function
                          F=( // Helper function to run through all substrings in a[0]
                          l, // Length
                          n, // Start position
                          w=a[0].substr(n,l) // The substring
                          )=>
                          l? // If l > 0:
                          n<0? // If n < 0:
                          F(--l,L-l) // Check another length
                          :a.some( // If n >= 0:
                          y=>y.indexOf(w)<0 // Check whether there is any string not containing the substring
                          // (indexOf used because of presence of regex special characters)
                          )? // If so:
                          F(l,n-1) // Check another substring
                          :w // If not, return this substring and terminate
                          // (This function checks from the longest substring possible, so
                          // it is safe to return right here)
                          :"" // If l <= 0: Return empty string (no common substring)
                          )(
                          L=a[0].length, // Starts from length = the whole length of a[0]
                          0 // And start position = 0
                          )





                          share|improve this answer











                          $endgroup$













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                            23 Answers
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                            8












                            $begingroup$


                            Brachylog (v2), 3 9 bytes



                            {sᵛ}ᶠlᵒtw


                            Try it online!



                            Full program. Input from standard input (as a JSON-style list of strings), output to standard output.



                            Explanation



                            {sᵛ}ᶠlᵒtw
                            s Find a substring
                            ᵛ of every element {of the input}; the same one for each
                            { }ᶠ Convert generator to list
                            lᵒt Take list element with maximum length
                            w Output it


                            Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)



                            Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.






                            share|improve this answer











                            $endgroup$









                            • 1




                              $begingroup$
                              I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                              $endgroup$
                              – t-clausen.dk
                              18 hours ago










                            • $begingroup$
                              Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                              $endgroup$
                              – ais523
                              18 hours ago












                            • $begingroup$
                              @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                              $endgroup$
                              – Kroppeb
                              13 hours ago
















                            8












                            $begingroup$


                            Brachylog (v2), 3 9 bytes



                            {sᵛ}ᶠlᵒtw


                            Try it online!



                            Full program. Input from standard input (as a JSON-style list of strings), output to standard output.



                            Explanation



                            {sᵛ}ᶠlᵒtw
                            s Find a substring
                            ᵛ of every element {of the input}; the same one for each
                            { }ᶠ Convert generator to list
                            lᵒt Take list element with maximum length
                            w Output it


                            Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)



                            Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.






                            share|improve this answer











                            $endgroup$









                            • 1




                              $begingroup$
                              I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                              $endgroup$
                              – t-clausen.dk
                              18 hours ago










                            • $begingroup$
                              Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                              $endgroup$
                              – ais523
                              18 hours ago












                            • $begingroup$
                              @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                              $endgroup$
                              – Kroppeb
                              13 hours ago














                            8












                            8








                            8





                            $begingroup$


                            Brachylog (v2), 3 9 bytes



                            {sᵛ}ᶠlᵒtw


                            Try it online!



                            Full program. Input from standard input (as a JSON-style list of strings), output to standard output.



                            Explanation



                            {sᵛ}ᶠlᵒtw
                            s Find a substring
                            ᵛ of every element {of the input}; the same one for each
                            { }ᶠ Convert generator to list
                            lᵒt Take list element with maximum length
                            w Output it


                            Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)



                            Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.






                            share|improve this answer











                            $endgroup$




                            Brachylog (v2), 3 9 bytes



                            {sᵛ}ᶠlᵒtw


                            Try it online!



                            Full program. Input from standard input (as a JSON-style list of strings), output to standard output.



                            Explanation



                            {sᵛ}ᶠlᵒtw
                            s Find a substring
                            ᵛ of every element {of the input}; the same one for each
                            { }ᶠ Convert generator to list
                            lᵒt Take list element with maximum length
                            w Output it


                            Apparently, the tiebreak order on s is not what it is in nearly everything else in Brachylog, so we need to manually override it to produce the longest output. (That's a bit frustrating: four extra characters for the override, plus two grouping characters because Brachylog doesn't parse two metapredicates in a row.)



                            Brachylog's s doesn't return empty substrings, so we need a bit of a trick to get around that: instead of making a function submission (which is what's normally done), we write a full program, outputting to standard output. That way, if there's a common substring, we just output it, and we're done. If there isn't a common substring, the program errors out – but it still prints nothing to standard output, thus it outputs the null string as intended.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited 18 hours ago


























                            community wiki





                            2 revs
                            ais523









                            • 1




                              $begingroup$
                              I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                              $endgroup$
                              – t-clausen.dk
                              18 hours ago










                            • $begingroup$
                              Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                              $endgroup$
                              – ais523
                              18 hours ago












                            • $begingroup$
                              @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                              $endgroup$
                              – Kroppeb
                              13 hours ago














                            • 1




                              $begingroup$
                              I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                              $endgroup$
                              – t-clausen.dk
                              18 hours ago










                            • $begingroup$
                              Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                              $endgroup$
                              – ais523
                              18 hours ago












                            • $begingroup$
                              @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                              $endgroup$
                              – Kroppeb
                              13 hours ago








                            1




                            1




                            $begingroup$
                            I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                            $endgroup$
                            – t-clausen.dk
                            18 hours ago




                            $begingroup$
                            I tried this with the input ["many inpuabts", "any inabputs", "ny iabii", "yanabny"]. I expected the results ab and ny. But only got the result a. Am I doing something wrong ?
                            $endgroup$
                            – t-clausen.dk
                            18 hours ago












                            $begingroup$
                            Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                            $endgroup$
                            – ais523
                            18 hours ago






                            $begingroup$
                            Ugh, seems I remembered the tiebreak order for s wrong, and overriding the tiebreak order is rather expensive byte-wise. Doing that now anyway, though, because it's important for the answer to be correct. Somehow none of the test cases I tried noticed the difference.
                            $endgroup$
                            – ais523
                            18 hours ago














                            $begingroup$
                            @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                            $endgroup$
                            – Kroppeb
                            13 hours ago




                            $begingroup$
                            @ais523 The order s produces substrings is by first giving all prefixes of the input (longest first), then dropping the first one and repeat
                            $endgroup$
                            – Kroppeb
                            13 hours ago











                            7












                            $begingroup$


                            Python 2, 82 bytes





                            f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)


                            Try it online!



                            Takes input splatted. Will time out for inputs where the first string is long.



                            The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.





                            Python 2, 94 bytes





                            lambda l:max(set.intersection(*map(g,l)),key=len)
                            g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}


                            Try it online!



                            A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.






                            share|improve this answer









                            $endgroup$


















                              7












                              $begingroup$


                              Python 2, 82 bytes





                              f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)


                              Try it online!



                              Takes input splatted. Will time out for inputs where the first string is long.



                              The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.





                              Python 2, 94 bytes





                              lambda l:max(set.intersection(*map(g,l)),key=len)
                              g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}


                              Try it online!



                              A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.






                              share|improve this answer









                              $endgroup$
















                                7












                                7








                                7





                                $begingroup$


                                Python 2, 82 bytes





                                f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)


                                Try it online!



                                Takes input splatted. Will time out for inputs where the first string is long.



                                The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.





                                Python 2, 94 bytes





                                lambda l:max(set.intersection(*map(g,l)),key=len)
                                g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}


                                Try it online!



                                A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.






                                share|improve this answer









                                $endgroup$




                                Python 2, 82 bytes





                                f=lambda h,*t:h and max(h*all(h in s for s in t),f(h[1:],*t),f(h[:-1],*t),key=len)


                                Try it online!



                                Takes input splatted. Will time out for inputs where the first string is long.



                                The idea is to take substrings of the first strings h to find the longest one that appears in all the remaining strings t. To do so, we recursively branch on removing the first or last character of h.





                                Python 2, 94 bytes





                                lambda l:max(set.intersection(*map(g,l)),key=len)
                                g=lambda s:s and{s}|g(s[1:])|g(s[:-1])or{''}


                                Try it online!



                                A more direct method. The auxiliary function g generates the set all substrings of s, and the main function takes the longest one in their intersection.







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered yesterday









                                xnorxnor

                                92.9k18190447




                                92.9k18190447























                                    4












                                    $begingroup$


                                    Ruby 2.6, 76 59 54 bytes





                                    ->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}


                                    Try it online! - Ruby 2.5 version (56 bytes)



                                    How?



                                    Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.



                                    Thanks Kirill L for -2 bytes and histocrat for another -2






                                    share|improve this answer











                                    $endgroup$









                                    • 1




                                      $begingroup$
                                      You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                      $endgroup$
                                      – Kirill L.
                                      14 hours ago
















                                    4












                                    $begingroup$


                                    Ruby 2.6, 76 59 54 bytes





                                    ->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}


                                    Try it online! - Ruby 2.5 version (56 bytes)



                                    How?



                                    Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.



                                    Thanks Kirill L for -2 bytes and histocrat for another -2






                                    share|improve this answer











                                    $endgroup$









                                    • 1




                                      $begingroup$
                                      You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                      $endgroup$
                                      – Kirill L.
                                      14 hours ago














                                    4












                                    4








                                    4





                                    $begingroup$


                                    Ruby 2.6, 76 59 54 bytes





                                    ->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}


                                    Try it online! - Ruby 2.5 version (56 bytes)



                                    How?



                                    Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.



                                    Thanks Kirill L for -2 bytes and histocrat for another -2






                                    share|improve this answer











                                    $endgroup$




                                    Ruby 2.6, 76 59 54 bytes





                                    ->a{*w=a;w.find{|r|w<<r.chop<<r[1..];a.all?{|s|s[r]}}}


                                    Try it online! - Ruby 2.5 version (56 bytes)



                                    How?



                                    Create a list of potential matches, initially set to the original array. Iterate on the list, and if a string does not match, add 2 new strings to the tail of the list, chopping off the first or the last character. At the end a match (eventually an empty string) will be found.



                                    Thanks Kirill L for -2 bytes and histocrat for another -2







                                    share|improve this answer














                                    share|improve this answer



                                    share|improve this answer








                                    edited 11 hours ago

























                                    answered 22 hours ago









                                    G BG B

                                    8,0561429




                                    8,0561429








                                    • 1




                                      $begingroup$
                                      You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                      $endgroup$
                                      – Kirill L.
                                      14 hours ago














                                    • 1




                                      $begingroup$
                                      You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                      $endgroup$
                                      – Kirill L.
                                      14 hours ago








                                    1




                                    1




                                    $begingroup$
                                    You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                    $endgroup$
                                    – Kirill L.
                                    14 hours ago




                                    $begingroup$
                                    You might as well say that it is Ruby 2.6 and get rid of -1 :)
                                    $endgroup$
                                    – Kirill L.
                                    14 hours ago











                                    4












                                    $begingroup$


                                    Zsh, 126 123 bytes



                                    3 bytes saved via arithmetic for



                                    for l in $@
                                    {a= i=
                                    for ((;i++<$#l**2;)){a+=($l[1+i/$#l,1+i%$#l])}
                                    b=(${${b-$a}:*a})}
                                    for s in $b;{(($#x<$#s))&&x=$s;}
                                    <<<$x


                                    Try it online!



                                    We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.



                                    for l in $@; {
                                    a= # empty a
                                    for (( ; i++ < $#l**2; )) { # compound double loop using div/mod
                                    a+=( $l[1+i/$#l,1+i%$#l] ) # append to a all possible substrings of the given line
                                    }
                                    b=( ${${b-$a}:*a} )
                                    # ${b-$a} # if b is unset substitute $a
                                    # ${ :*a} # take common elements of ${b-$a} and $a
                                    # b=( ) # set b to those elements
                                    }
                                    for s in $b; { # for every common substring
                                    (( $#x < $#s )) && x=$s # if the current word is longer, use it
                                    }
                                    <<<$x # print to stdout





                                    share|improve this answer










                                    New contributor




                                    GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                    Check out our Code of Conduct.






                                    $endgroup$


















                                      4












                                      $begingroup$


                                      Zsh, 126 123 bytes



                                      3 bytes saved via arithmetic for



                                      for l in $@
                                      {a= i=
                                      for ((;i++<$#l**2;)){a+=($l[1+i/$#l,1+i%$#l])}
                                      b=(${${b-$a}:*a})}
                                      for s in $b;{(($#x<$#s))&&x=$s;}
                                      <<<$x


                                      Try it online!



                                      We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.



                                      for l in $@; {
                                      a= # empty a
                                      for (( ; i++ < $#l**2; )) { # compound double loop using div/mod
                                      a+=( $l[1+i/$#l,1+i%$#l] ) # append to a all possible substrings of the given line
                                      }
                                      b=( ${${b-$a}:*a} )
                                      # ${b-$a} # if b is unset substitute $a
                                      # ${ :*a} # take common elements of ${b-$a} and $a
                                      # b=( ) # set b to those elements
                                      }
                                      for s in $b; { # for every common substring
                                      (( $#x < $#s )) && x=$s # if the current word is longer, use it
                                      }
                                      <<<$x # print to stdout





                                      share|improve this answer










                                      New contributor




                                      GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                      Check out our Code of Conduct.






                                      $endgroup$
















                                        4












                                        4








                                        4





                                        $begingroup$


                                        Zsh, 126 123 bytes



                                        3 bytes saved via arithmetic for



                                        for l in $@
                                        {a= i=
                                        for ((;i++<$#l**2;)){a+=($l[1+i/$#l,1+i%$#l])}
                                        b=(${${b-$a}:*a})}
                                        for s in $b;{(($#x<$#s))&&x=$s;}
                                        <<<$x


                                        Try it online!



                                        We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.



                                        for l in $@; {
                                        a= # empty a
                                        for (( ; i++ < $#l**2; )) { # compound double loop using div/mod
                                        a+=( $l[1+i/$#l,1+i%$#l] ) # append to a all possible substrings of the given line
                                        }
                                        b=( ${${b-$a}:*a} )
                                        # ${b-$a} # if b is unset substitute $a
                                        # ${ :*a} # take common elements of ${b-$a} and $a
                                        # b=( ) # set b to those elements
                                        }
                                        for s in $b; { # for every common substring
                                        (( $#x < $#s )) && x=$s # if the current word is longer, use it
                                        }
                                        <<<$x # print to stdout





                                        share|improve this answer










                                        New contributor




                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        $endgroup$




                                        Zsh, 126 123 bytes



                                        3 bytes saved via arithmetic for



                                        for l in $@
                                        {a= i=
                                        for ((;i++<$#l**2;)){a+=($l[1+i/$#l,1+i%$#l])}
                                        b=(${${b-$a}:*a})}
                                        for s in $b;{(($#x<$#s))&&x=$s;}
                                        <<<$x


                                        Try it online!



                                        We read all possible substrings into the arraya, and then set b to the intersection of the arrays a and b. The construct ${b-$a} will only substitue $a on the first iteration: Unlike its sibling expansion ${b:-$a}, it will not substitute when b is set but empty.



                                        for l in $@; {
                                        a= # empty a
                                        for (( ; i++ < $#l**2; )) { # compound double loop using div/mod
                                        a+=( $l[1+i/$#l,1+i%$#l] ) # append to a all possible substrings of the given line
                                        }
                                        b=( ${${b-$a}:*a} )
                                        # ${b-$a} # if b is unset substitute $a
                                        # ${ :*a} # take common elements of ${b-$a} and $a
                                        # b=( ) # set b to those elements
                                        }
                                        for s in $b; { # for every common substring
                                        (( $#x < $#s )) && x=$s # if the current word is longer, use it
                                        }
                                        <<<$x # print to stdout






                                        share|improve this answer










                                        New contributor




                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        share|improve this answer



                                        share|improve this answer








                                        edited 1 hour ago





















                                        New contributor




                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.









                                        answered yesterday









                                        GammaFunctionGammaFunction

                                        1315




                                        1315




                                        New contributor




                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.





                                        New contributor





                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.






                                        GammaFunction is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                        Check out our Code of Conduct.























                                            3












                                            $begingroup$


                                            Perl 6, 62 bytes





                                            {~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}


                                            Try it online!



                                            I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.






                                            share|improve this answer









                                            $endgroup$









                                            • 3




                                              $begingroup$
                                              max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                              $endgroup$
                                              – Ven
                                              21 hours ago


















                                            3












                                            $begingroup$


                                            Perl 6, 62 bytes





                                            {~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}


                                            Try it online!



                                            I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.






                                            share|improve this answer









                                            $endgroup$









                                            • 3




                                              $begingroup$
                                              max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                              $endgroup$
                                              – Ven
                                              21 hours ago
















                                            3












                                            3








                                            3





                                            $begingroup$


                                            Perl 6, 62 bytes





                                            {~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}


                                            Try it online!



                                            I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.






                                            share|improve this answer









                                            $endgroup$




                                            Perl 6, 62 bytes





                                            {~sort(-*.comb,keys [∩] .map(*.comb[^*X.. ^*+1]>>.join))[0]}


                                            Try it online!



                                            I'm a little annoyed the Perl 6 can't do set operations on lists of lists, which is why there's an extra .comb and >> in there. Another annoying thing is that max can't take an function for how to compare items, meaning I have to user sort instead.







                                            share|improve this answer












                                            share|improve this answer



                                            share|improve this answer










                                            answered yesterday









                                            Jo KingJo King

                                            25.5k362129




                                            25.5k362129








                                            • 3




                                              $begingroup$
                                              max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                              $endgroup$
                                              – Ven
                                              21 hours ago
















                                            • 3




                                              $begingroup$
                                              max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                              $endgroup$
                                              – Ven
                                              21 hours ago










                                            3




                                            3




                                            $begingroup$
                                            max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                            $endgroup$
                                            – Ven
                                            21 hours ago






                                            $begingroup$
                                            max can take such a function (arity 1 though), either by position when called in OO mode, or a named :by argument in procedural mode.
                                            $endgroup$
                                            – Ven
                                            21 hours ago













                                            3












                                            $begingroup$


                                            05AB1E, 14 9 bytes



                                            €Œ.«Ãõªéθ


                                            -5 bytes thanks to @Adnan.



                                            Try it online or verify all test cases.



                                            Explanation:





                                            €Œ         # Get the substring of each string in the (implicit) input-list
                                            .« # Right-reduce this list of list of strings by:
                                            Ã # Only keep all the strings that are present in both list of strings
                                            õª # Append an empty string to the remaining list of strings
                                            é # Sort by length
                                            θ # And pop and push its last item (which is output implicitly as result)





                                            share|improve this answer











                                            $endgroup$









                                            • 1




                                              $begingroup$
                                              I think €Œ.«Ãõªéθ should work for 9 bytes.
                                              $endgroup$
                                              – Adnan
                                              16 hours ago










                                            • $begingroup$
                                              @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              16 hours ago










                                            • $begingroup$
                                              Actually, I think €Œ.«ÃéθJ should work for 8.
                                              $endgroup$
                                              – Adnan
                                              10 hours ago
















                                            3












                                            $begingroup$


                                            05AB1E, 14 9 bytes



                                            €Œ.«Ãõªéθ


                                            -5 bytes thanks to @Adnan.



                                            Try it online or verify all test cases.



                                            Explanation:





                                            €Œ         # Get the substring of each string in the (implicit) input-list
                                            .« # Right-reduce this list of list of strings by:
                                            Ã # Only keep all the strings that are present in both list of strings
                                            õª # Append an empty string to the remaining list of strings
                                            é # Sort by length
                                            θ # And pop and push its last item (which is output implicitly as result)





                                            share|improve this answer











                                            $endgroup$









                                            • 1




                                              $begingroup$
                                              I think €Œ.«Ãõªéθ should work for 9 bytes.
                                              $endgroup$
                                              – Adnan
                                              16 hours ago










                                            • $begingroup$
                                              @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              16 hours ago










                                            • $begingroup$
                                              Actually, I think €Œ.«ÃéθJ should work for 8.
                                              $endgroup$
                                              – Adnan
                                              10 hours ago














                                            3












                                            3








                                            3





                                            $begingroup$


                                            05AB1E, 14 9 bytes



                                            €Œ.«Ãõªéθ


                                            -5 bytes thanks to @Adnan.



                                            Try it online or verify all test cases.



                                            Explanation:





                                            €Œ         # Get the substring of each string in the (implicit) input-list
                                            .« # Right-reduce this list of list of strings by:
                                            Ã # Only keep all the strings that are present in both list of strings
                                            õª # Append an empty string to the remaining list of strings
                                            é # Sort by length
                                            θ # And pop and push its last item (which is output implicitly as result)





                                            share|improve this answer











                                            $endgroup$




                                            05AB1E, 14 9 bytes



                                            €Œ.«Ãõªéθ


                                            -5 bytes thanks to @Adnan.



                                            Try it online or verify all test cases.



                                            Explanation:





                                            €Œ         # Get the substring of each string in the (implicit) input-list
                                            .« # Right-reduce this list of list of strings by:
                                            Ã # Only keep all the strings that are present in both list of strings
                                            õª # Append an empty string to the remaining list of strings
                                            é # Sort by length
                                            θ # And pop and push its last item (which is output implicitly as result)






                                            share|improve this answer














                                            share|improve this answer



                                            share|improve this answer








                                            edited 16 hours ago

























                                            answered 19 hours ago









                                            Kevin CruijssenKevin Cruijssen

                                            41.3k567213




                                            41.3k567213








                                            • 1




                                              $begingroup$
                                              I think €Œ.«Ãõªéθ should work for 9 bytes.
                                              $endgroup$
                                              – Adnan
                                              16 hours ago










                                            • $begingroup$
                                              @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              16 hours ago










                                            • $begingroup$
                                              Actually, I think €Œ.«ÃéθJ should work for 8.
                                              $endgroup$
                                              – Adnan
                                              10 hours ago














                                            • 1




                                              $begingroup$
                                              I think €Œ.«Ãõªéθ should work for 9 bytes.
                                              $endgroup$
                                              – Adnan
                                              16 hours ago










                                            • $begingroup$
                                              @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                              $endgroup$
                                              – Kevin Cruijssen
                                              16 hours ago










                                            • $begingroup$
                                              Actually, I think €Œ.«ÃéθJ should work for 8.
                                              $endgroup$
                                              – Adnan
                                              10 hours ago








                                            1




                                            1




                                            $begingroup$
                                            I think €Œ.«Ãõªéθ should work for 9 bytes.
                                            $endgroup$
                                            – Adnan
                                            16 hours ago




                                            $begingroup$
                                            I think €Œ.«Ãõªéθ should work for 9 bytes.
                                            $endgroup$
                                            – Adnan
                                            16 hours ago












                                            $begingroup$
                                            @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                            $endgroup$
                                            – Kevin Cruijssen
                                            16 hours ago




                                            $begingroup$
                                            @Adnan Wait, we do have a reduce.. How did I miss that. :S I tried Å«Ã, but didn't realize I should have used .«Ã instead.. Thanks!
                                            $endgroup$
                                            – Kevin Cruijssen
                                            16 hours ago












                                            $begingroup$
                                            Actually, I think €Œ.«ÃéθJ should work for 8.
                                            $endgroup$
                                            – Adnan
                                            10 hours ago




                                            $begingroup$
                                            Actually, I think €Œ.«ÃéθJ should work for 8.
                                            $endgroup$
                                            – Adnan
                                            10 hours ago











                                            3












                                            $begingroup$


                                            Jelly, 12 6 bytes



                                            Ẇ€f/ṫ0


                                            Try it online!



                                            Thanks to @JonathanAllan for saving 6 bytes!






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              also I think œ&/ may be replaced by f/ here saving another
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago












                                            • $begingroup$
                                              @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                              $endgroup$
                                              – Nick Kennedy
                                              14 hours ago










                                            • $begingroup$
                                              Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago
















                                            3












                                            $begingroup$


                                            Jelly, 12 6 bytes



                                            Ẇ€f/ṫ0


                                            Try it online!



                                            Thanks to @JonathanAllan for saving 6 bytes!






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              also I think œ&/ may be replaced by f/ here saving another
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago












                                            • $begingroup$
                                              @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                              $endgroup$
                                              – Nick Kennedy
                                              14 hours ago










                                            • $begingroup$
                                              Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago














                                            3












                                            3








                                            3





                                            $begingroup$


                                            Jelly, 12 6 bytes



                                            Ẇ€f/ṫ0


                                            Try it online!



                                            Thanks to @JonathanAllan for saving 6 bytes!






                                            share|improve this answer











                                            $endgroup$




                                            Jelly, 12 6 bytes



                                            Ẇ€f/ṫ0


                                            Try it online!



                                            Thanks to @JonathanAllan for saving 6 bytes!







                                            share|improve this answer














                                            share|improve this answer



                                            share|improve this answer








                                            edited 12 hours ago

























                                            answered yesterday









                                            Nick KennedyNick Kennedy

                                            99647




                                            99647












                                            • $begingroup$
                                              I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              also I think œ&/ may be replaced by f/ here saving another
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago












                                            • $begingroup$
                                              @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                              $endgroup$
                                              – Nick Kennedy
                                              14 hours ago










                                            • $begingroup$
                                              Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago


















                                            • $begingroup$
                                              I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              also I think œ&/ may be replaced by f/ here saving another
                                              $endgroup$
                                              – Jonathan Allan
                                              18 hours ago










                                            • $begingroup$
                                              I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago












                                            • $begingroup$
                                              @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                              $endgroup$
                                              – Nick Kennedy
                                              14 hours ago










                                            • $begingroup$
                                              Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                              $endgroup$
                                              – Jonathan Allan
                                              14 hours ago
















                                            $begingroup$
                                            I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                            $endgroup$
                                            – Jonathan Allan
                                            18 hours ago




                                            $begingroup$
                                            I believe Ẇ€œ&/Ṫḟ0 would do the job and save four bytes since the sub-strings are already ordered by length, hence the filtered result will be; then all that remains is that when there are no matches tail produces a zero, and since we are guaranteed lists of characters we can simply filter those out.
                                            $endgroup$
                                            – Jonathan Allan
                                            18 hours ago












                                            $begingroup$
                                            also I think œ&/ may be replaced by f/ here saving another
                                            $endgroup$
                                            – Jonathan Allan
                                            18 hours ago




                                            $begingroup$
                                            also I think œ&/ may be replaced by f/ here saving another
                                            $endgroup$
                                            – Jonathan Allan
                                            18 hours ago












                                            $begingroup$
                                            I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                            $endgroup$
                                            – Jonathan Allan
                                            14 hours ago






                                            $begingroup$
                                            I submitted a pull request to (hopefully) make the result of reducing an empty list be an empty list (rather than raising a TypeError). If that gets merged in I believe that this answer could then become a six-byter with Ẇ€f/ṛ/.
                                            $endgroup$
                                            – Jonathan Allan
                                            14 hours ago














                                            $begingroup$
                                            @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                            $endgroup$
                                            – Nick Kennedy
                                            14 hours ago




                                            $begingroup$
                                            @JonathanAllan sounds good. Thanks for the other tips - hope you were happy for me to incorporate them.
                                            $endgroup$
                                            – Nick Kennedy
                                            14 hours ago












                                            $begingroup$
                                            Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                            $endgroup$
                                            – Jonathan Allan
                                            14 hours ago




                                            $begingroup$
                                            Yes, my reason for those comments was to allow you to incorporate the ideas into your post.
                                            $endgroup$
                                            – Jonathan Allan
                                            14 hours ago











                                            3












                                            $begingroup$


                                            R, 119 116 108 106 bytes





                                            function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]


                                            Try it online!



                                            Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.



                                            -3 bytes thanks to Kirill L.



                                            -8 bytes using lapply instead of Map



                                            -2 bytes thanks to Kirill L. again, removing braces






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                              $endgroup$
                                              – Kirill L.
                                              14 hours ago










                                            • $begingroup$
                                              @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                              $endgroup$
                                              – Giuseppe
                                              14 hours ago












                                            • $begingroup$
                                              You can also drop braces for another -2
                                              $endgroup$
                                              – Kirill L.
                                              8 hours ago










                                            • $begingroup$
                                              @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                              $endgroup$
                                              – Giuseppe
                                              8 hours ago










                                            • $begingroup$
                                              that's the problem with golfing a language you use every day
                                              $endgroup$
                                              – MickyT
                                              8 hours ago
















                                            3












                                            $begingroup$


                                            R, 119 116 108 106 bytes





                                            function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]


                                            Try it online!



                                            Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.



                                            -3 bytes thanks to Kirill L.



                                            -8 bytes using lapply instead of Map



                                            -2 bytes thanks to Kirill L. again, removing braces






                                            share|improve this answer











                                            $endgroup$













                                            • $begingroup$
                                              I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                              $endgroup$
                                              – Kirill L.
                                              14 hours ago










                                            • $begingroup$
                                              @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                              $endgroup$
                                              – Giuseppe
                                              14 hours ago












                                            • $begingroup$
                                              You can also drop braces for another -2
                                              $endgroup$
                                              – Kirill L.
                                              8 hours ago










                                            • $begingroup$
                                              @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                              $endgroup$
                                              – Giuseppe
                                              8 hours ago










                                            • $begingroup$
                                              that's the problem with golfing a language you use every day
                                              $endgroup$
                                              – MickyT
                                              8 hours ago














                                            3












                                            3








                                            3





                                            $begingroup$


                                            R, 119 116 108 106 bytes





                                            function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]


                                            Try it online!



                                            Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.



                                            -3 bytes thanks to Kirill L.



                                            -8 bytes using lapply instead of Map



                                            -2 bytes thanks to Kirill L. again, removing braces






                                            share|improve this answer











                                            $endgroup$




                                            R, 119 116 108 106 bytes





                                            function(S,`?`=nchar,K=max(?S),s=Reduce(intersect,lapply(S,substring,0:K,rep(0:K,e=K+1))))s[which.max(?s)]


                                            Try it online!



                                            Find all substrings of each string, find the intersection of each list of substrings, then finally return (one of) the longest.



                                            -3 bytes thanks to Kirill L.



                                            -8 bytes using lapply instead of Map



                                            -2 bytes thanks to Kirill L. again, removing braces







                                            share|improve this answer














                                            share|improve this answer



                                            share|improve this answer








                                            edited 8 hours ago

























                                            answered 14 hours ago









                                            GiuseppeGiuseppe

                                            17.1k31152




                                            17.1k31152












                                            • $begingroup$
                                              I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                              $endgroup$
                                              – Kirill L.
                                              14 hours ago










                                            • $begingroup$
                                              @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                              $endgroup$
                                              – Giuseppe
                                              14 hours ago












                                            • $begingroup$
                                              You can also drop braces for another -2
                                              $endgroup$
                                              – Kirill L.
                                              8 hours ago










                                            • $begingroup$
                                              @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                              $endgroup$
                                              – Giuseppe
                                              8 hours ago










                                            • $begingroup$
                                              that's the problem with golfing a language you use every day
                                              $endgroup$
                                              – MickyT
                                              8 hours ago


















                                            • $begingroup$
                                              I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                              $endgroup$
                                              – Kirill L.
                                              14 hours ago










                                            • $begingroup$
                                              @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                              $endgroup$
                                              – Giuseppe
                                              14 hours ago












                                            • $begingroup$
                                              You can also drop braces for another -2
                                              $endgroup$
                                              – Kirill L.
                                              8 hours ago










                                            • $begingroup$
                                              @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                              $endgroup$
                                              – Giuseppe
                                              8 hours ago










                                            • $begingroup$
                                              that's the problem with golfing a language you use every day
                                              $endgroup$
                                              – MickyT
                                              8 hours ago
















                                            $begingroup$
                                            I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                            $endgroup$
                                            – Kirill L.
                                            14 hours ago




                                            $begingroup$
                                            I don't have time to check, but if I'm not mistaken, 2 occurrences of nchar are enough to save something by declaring nchar as an unary operator.
                                            $endgroup$
                                            – Kirill L.
                                            14 hours ago












                                            $begingroup$
                                            @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                            $endgroup$
                                            – Giuseppe
                                            14 hours ago






                                            $begingroup$
                                            @KirillL. yes, it is shorter by 2 bytes. Thanks! Aliasing list similarly gives us -3 bytes.
                                            $endgroup$
                                            – Giuseppe
                                            14 hours ago














                                            $begingroup$
                                            You can also drop braces for another -2
                                            $endgroup$
                                            – Kirill L.
                                            8 hours ago




                                            $begingroup$
                                            You can also drop braces for another -2
                                            $endgroup$
                                            – Kirill L.
                                            8 hours ago












                                            $begingroup$
                                            @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                            $endgroup$
                                            – Giuseppe
                                            8 hours ago




                                            $begingroup$
                                            @KirillL. thanks! I'm a bit worried I'm going to start doing that with production code...
                                            $endgroup$
                                            – Giuseppe
                                            8 hours ago












                                            $begingroup$
                                            that's the problem with golfing a language you use every day
                                            $endgroup$
                                            – MickyT
                                            8 hours ago




                                            $begingroup$
                                            that's the problem with golfing a language you use every day
                                            $endgroup$
                                            – MickyT
                                            8 hours ago











                                            2












                                            $begingroup$


                                            Haskell, 80 bytes





                                            import Data.List
                                            f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]


                                            Try it online!



                                            Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.






                                            share|improve this answer











                                            $endgroup$


















                                              2












                                              $begingroup$


                                              Haskell, 80 bytes





                                              import Data.List
                                              f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]


                                              Try it online!



                                              Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.






                                              share|improve this answer











                                              $endgroup$
















                                                2












                                                2








                                                2





                                                $begingroup$


                                                Haskell, 80 bytes





                                                import Data.List
                                                f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]


                                                Try it online!



                                                Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.






                                                share|improve this answer











                                                $endgroup$




                                                Haskell, 80 bytes





                                                import Data.List
                                                f(x:r)=last$sortOn(0<$)[s|s<-inits=<<tails x,all(isInfixOf s)r]


                                                Try it online!



                                                Get all suffixes (tails) of the first word x in the list and take all prefixes (inits) of those suffixes to get all substrings s of x. Keep each s that isInfixOf all strings in the remaining list r. Sort those substrings by length (using the (0<$) trick) and return the last.







                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited 22 hours ago

























                                                answered 22 hours ago









                                                LaikoniLaikoni

                                                20.3k439103




                                                20.3k439103























                                                    2












                                                    $begingroup$

                                                    C#, 320 bytes



                                                    //method that returns all possible substrings from a string
                                                    IEnumerable<string>f(string s){for(int i=0;i<s.Length;i++)for(int j=i;j<s.Length;j++)yield return s.Substring(i,j-i+1);}

                                                    //main method
                                                    void g(List<string>l){Console.WriteLine(string.Join(",",l.Select(s=>f(s)).Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x=>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));}


                                                    Try it online!



                                                    Couldn't make it work for https://tio.run/#cs-csc, would be really grateful if someone showed me how to input lists on that site!



                                                    The result can be outputted as a collection rather than a string, which would save a few bytes, but it seems like in C#, unlike with other languages, it won't be readable through Console.Writeline, would that still be a valid answer or is there a different workaround?






                                                    share|improve this answer








                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$













                                                    • $begingroup$
                                                      You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      14 hours ago












                                                    • $begingroup$
                                                      Here's the solution golfed down some bytes 215 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      13 hours ago
















                                                    2












                                                    $begingroup$

                                                    C#, 320 bytes



                                                    //method that returns all possible substrings from a string
                                                    IEnumerable<string>f(string s){for(int i=0;i<s.Length;i++)for(int j=i;j<s.Length;j++)yield return s.Substring(i,j-i+1);}

                                                    //main method
                                                    void g(List<string>l){Console.WriteLine(string.Join(",",l.Select(s=>f(s)).Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x=>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));}


                                                    Try it online!



                                                    Couldn't make it work for https://tio.run/#cs-csc, would be really grateful if someone showed me how to input lists on that site!



                                                    The result can be outputted as a collection rather than a string, which would save a few bytes, but it seems like in C#, unlike with other languages, it won't be readable through Console.Writeline, would that still be a valid answer or is there a different workaround?






                                                    share|improve this answer








                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$













                                                    • $begingroup$
                                                      You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      14 hours ago












                                                    • $begingroup$
                                                      Here's the solution golfed down some bytes 215 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      13 hours ago














                                                    2












                                                    2








                                                    2





                                                    $begingroup$

                                                    C#, 320 bytes



                                                    //method that returns all possible substrings from a string
                                                    IEnumerable<string>f(string s){for(int i=0;i<s.Length;i++)for(int j=i;j<s.Length;j++)yield return s.Substring(i,j-i+1);}

                                                    //main method
                                                    void g(List<string>l){Console.WriteLine(string.Join(",",l.Select(s=>f(s)).Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x=>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));}


                                                    Try it online!



                                                    Couldn't make it work for https://tio.run/#cs-csc, would be really grateful if someone showed me how to input lists on that site!



                                                    The result can be outputted as a collection rather than a string, which would save a few bytes, but it seems like in C#, unlike with other languages, it won't be readable through Console.Writeline, would that still be a valid answer or is there a different workaround?






                                                    share|improve this answer








                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$



                                                    C#, 320 bytes



                                                    //method that returns all possible substrings from a string
                                                    IEnumerable<string>f(string s){for(int i=0;i<s.Length;i++)for(int j=i;j<s.Length;j++)yield return s.Substring(i,j-i+1);}

                                                    //main method
                                                    void g(List<string>l){Console.WriteLine(string.Join(",",l.Select(s=>f(s)).Aggregate((a,b)=>a.Intersect(b)).GroupBy(x=>x.Length).OrderBy(x=>x.Key).LastOrDefault()?.Select(y=>y)??new List<string>()));}


                                                    Try it online!



                                                    Couldn't make it work for https://tio.run/#cs-csc, would be really grateful if someone showed me how to input lists on that site!



                                                    The result can be outputted as a collection rather than a string, which would save a few bytes, but it seems like in C#, unlike with other languages, it won't be readable through Console.Writeline, would that still be a valid answer or is there a different workaround?







                                                    share|improve this answer








                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    share|improve this answer



                                                    share|improve this answer






                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    answered 19 hours ago









                                                    Innat3Innat3

                                                    1213




                                                    1213




                                                    New contributor




                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.





                                                    New contributor





                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    Innat3 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.












                                                    • $begingroup$
                                                      You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      14 hours ago












                                                    • $begingroup$
                                                      Here's the solution golfed down some bytes 215 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      13 hours ago


















                                                    • $begingroup$
                                                      You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      14 hours ago












                                                    • $begingroup$
                                                      Here's the solution golfed down some bytes 215 bytes
                                                      $endgroup$
                                                      – Expired Data
                                                      13 hours ago
















                                                    $begingroup$
                                                    You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                    $endgroup$
                                                    – Expired Data
                                                    14 hours ago






                                                    $begingroup$
                                                    You can just return the string from a function. So in Interactive compiler it might look something like this: 294 bytes
                                                    $endgroup$
                                                    – Expired Data
                                                    14 hours ago














                                                    $begingroup$
                                                    Here's the solution golfed down some bytes 215 bytes
                                                    $endgroup$
                                                    – Expired Data
                                                    13 hours ago




                                                    $begingroup$
                                                    Here's the solution golfed down some bytes 215 bytes
                                                    $endgroup$
                                                    – Expired Data
                                                    13 hours ago











                                                    2












                                                    $begingroup$


                                                    Japt v2.0a0 -hF, 8 bytes



                                                    Îã f@eøX


                                                    Thanks to Shaggy for saving 3 bytes



                                                    Try it



                                                    Îã              //Generate all substrings of the first string
                                                    f@ //Filter; keep the substrings that satisfy the following predicate:
                                                    e // If all strings of the input...
                                                    øX // Contain this substring, then keep it
                                                    -h //Take last element
                                                    -F //If last element is undefined, default to empty string





                                                    share|improve this answer











                                                    $endgroup$













                                                    • $begingroup$
                                                      You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                      $endgroup$
                                                      – Shaggy
                                                      19 hours ago
















                                                    2












                                                    $begingroup$


                                                    Japt v2.0a0 -hF, 8 bytes



                                                    Îã f@eøX


                                                    Thanks to Shaggy for saving 3 bytes



                                                    Try it



                                                    Îã              //Generate all substrings of the first string
                                                    f@ //Filter; keep the substrings that satisfy the following predicate:
                                                    e // If all strings of the input...
                                                    øX // Contain this substring, then keep it
                                                    -h //Take last element
                                                    -F //If last element is undefined, default to empty string





                                                    share|improve this answer











                                                    $endgroup$













                                                    • $begingroup$
                                                      You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                      $endgroup$
                                                      – Shaggy
                                                      19 hours ago














                                                    2












                                                    2








                                                    2





                                                    $begingroup$


                                                    Japt v2.0a0 -hF, 8 bytes



                                                    Îã f@eøX


                                                    Thanks to Shaggy for saving 3 bytes



                                                    Try it



                                                    Îã              //Generate all substrings of the first string
                                                    f@ //Filter; keep the substrings that satisfy the following predicate:
                                                    e // If all strings of the input...
                                                    øX // Contain this substring, then keep it
                                                    -h //Take last element
                                                    -F //If last element is undefined, default to empty string





                                                    share|improve this answer











                                                    $endgroup$




                                                    Japt v2.0a0 -hF, 8 bytes



                                                    Îã f@eøX


                                                    Thanks to Shaggy for saving 3 bytes



                                                    Try it



                                                    Îã              //Generate all substrings of the first string
                                                    f@ //Filter; keep the substrings that satisfy the following predicate:
                                                    e // If all strings of the input...
                                                    øX // Contain this substring, then keep it
                                                    -h //Take last element
                                                    -F //If last element is undefined, default to empty string






                                                    share|improve this answer














                                                    share|improve this answer



                                                    share|improve this answer








                                                    edited 12 hours ago

























                                                    answered yesterday









                                                    Embodiment of IgnoranceEmbodiment of Ignorance

                                                    2,168125




                                                    2,168125












                                                    • $begingroup$
                                                      You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                      $endgroup$
                                                      – Shaggy
                                                      19 hours ago


















                                                    • $begingroup$
                                                      You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                      $endgroup$
                                                      – Shaggy
                                                      19 hours ago
















                                                    $begingroup$
                                                    You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                    $endgroup$
                                                    – Shaggy
                                                    19 hours ago




                                                    $begingroup$
                                                    You shouldn't need to sort by length at the end, saving 3 bytes. Also, -F defaults to the empty string.
                                                    $endgroup$
                                                    – Shaggy
                                                    19 hours ago











                                                    1












                                                    $begingroup$


                                                    Python 3, 137 bytes





                                                    def a(b):c=[[d[f:e]for e in range(len(d)+1)for f in range(e+1)]for d in b];return max([i for i in c[0]if all(i in j for j in c)],key=len)


                                                    Try it online!






                                                    share|improve this answer










                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$













                                                    • $begingroup$
                                                      You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                      $endgroup$
                                                      – Shieru Asakoto
                                                      yesterday










                                                    • $begingroup$
                                                      @JoKing tio.run/…
                                                      $endgroup$
                                                      – Artemis Fowl
                                                      yesterday










                                                    • $begingroup$
                                                      Right, here's a fixed version of the 135 byte program
                                                      $endgroup$
                                                      – Jo King
                                                      yesterday


















                                                    1












                                                    $begingroup$


                                                    Python 3, 137 bytes





                                                    def a(b):c=[[d[f:e]for e in range(len(d)+1)for f in range(e+1)]for d in b];return max([i for i in c[0]if all(i in j for j in c)],key=len)


                                                    Try it online!






                                                    share|improve this answer










                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$













                                                    • $begingroup$
                                                      You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                      $endgroup$
                                                      – Shieru Asakoto
                                                      yesterday










                                                    • $begingroup$
                                                      @JoKing tio.run/…
                                                      $endgroup$
                                                      – Artemis Fowl
                                                      yesterday










                                                    • $begingroup$
                                                      Right, here's a fixed version of the 135 byte program
                                                      $endgroup$
                                                      – Jo King
                                                      yesterday
















                                                    1












                                                    1








                                                    1





                                                    $begingroup$


                                                    Python 3, 137 bytes





                                                    def a(b):c=[[d[f:e]for e in range(len(d)+1)for f in range(e+1)]for d in b];return max([i for i in c[0]if all(i in j for j in c)],key=len)


                                                    Try it online!






                                                    share|improve this answer










                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    $endgroup$




                                                    Python 3, 137 bytes





                                                    def a(b):c=[[d[f:e]for e in range(len(d)+1)for f in range(e+1)]for d in b];return max([i for i in c[0]if all(i in j for j in c)],key=len)


                                                    Try it online!







                                                    share|improve this answer










                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    share|improve this answer



                                                    share|improve this answer








                                                    edited yesterday





















                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.









                                                    answered yesterday









                                                    Artemis FowlArtemis Fowl

                                                    1114




                                                    1114




                                                    New contributor




                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.





                                                    New contributor





                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.






                                                    Artemis Fowl is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                    Check out our Code of Conduct.












                                                    • $begingroup$
                                                      You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                      $endgroup$
                                                      – Shieru Asakoto
                                                      yesterday










                                                    • $begingroup$
                                                      @JoKing tio.run/…
                                                      $endgroup$
                                                      – Artemis Fowl
                                                      yesterday










                                                    • $begingroup$
                                                      Right, here's a fixed version of the 135 byte program
                                                      $endgroup$
                                                      – Jo King
                                                      yesterday




















                                                    • $begingroup$
                                                      You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                      $endgroup$
                                                      – Shieru Asakoto
                                                      yesterday










                                                    • $begingroup$
                                                      @JoKing tio.run/…
                                                      $endgroup$
                                                      – Artemis Fowl
                                                      yesterday










                                                    • $begingroup$
                                                      Right, here's a fixed version of the 135 byte program
                                                      $endgroup$
                                                      – Jo King
                                                      yesterday


















                                                    $begingroup$
                                                    You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                    $endgroup$
                                                    – Shieru Asakoto
                                                    yesterday




                                                    $begingroup$
                                                    You may want to use single spaces as indentation instead of 4 that seems to shave more than 100 bytes.
                                                    $endgroup$
                                                    – Shieru Asakoto
                                                    yesterday












                                                    $begingroup$
                                                    @JoKing tio.run/…
                                                    $endgroup$
                                                    – Artemis Fowl
                                                    yesterday




                                                    $begingroup$
                                                    @JoKing tio.run/…
                                                    $endgroup$
                                                    – Artemis Fowl
                                                    yesterday












                                                    $begingroup$
                                                    Right, here's a fixed version of the 135 byte program
                                                    $endgroup$
                                                    – Jo King
                                                    yesterday






                                                    $begingroup$
                                                    Right, here's a fixed version of the 135 byte program
                                                    $endgroup$
                                                    – Jo King
                                                    yesterday













                                                    1












                                                    $begingroup$


                                                    Python 2, 103 bytes





                                                    lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)


                                                    Try it online!



                                                    This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.






                                                    share|improve this answer











                                                    $endgroup$













                                                    • $begingroup$
                                                      1 byte shorter with set.intersection.
                                                      $endgroup$
                                                      – ovs
                                                      19 hours ago
















                                                    1












                                                    $begingroup$


                                                    Python 2, 103 bytes





                                                    lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)


                                                    Try it online!



                                                    This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.






                                                    share|improve this answer











                                                    $endgroup$













                                                    • $begingroup$
                                                      1 byte shorter with set.intersection.
                                                      $endgroup$
                                                      – ovs
                                                      19 hours ago














                                                    1












                                                    1








                                                    1





                                                    $begingroup$


                                                    Python 2, 103 bytes





                                                    lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)


                                                    Try it online!



                                                    This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.






                                                    share|improve this answer











                                                    $endgroup$




                                                    Python 2, 103 bytes





                                                    lambda b:max(reduce(set.__and__,[{d[f:e]for e in range(len(d)+2)for f in range(e)}for d in b]),key=len)


                                                    Try it online!



                                                    This is an anonymous lambda that transforms each element into the set of all substrings, then reduces it by set intersection (set.__and__) and then returns the max element by length.







                                                    share|improve this answer














                                                    share|improve this answer



                                                    share|improve this answer








                                                    edited yesterday

























                                                    answered yesterday









                                                    Jo KingJo King

                                                    25.5k362129




                                                    25.5k362129












                                                    • $begingroup$
                                                      1 byte shorter with set.intersection.
                                                      $endgroup$
                                                      – ovs
                                                      19 hours ago


















                                                    • $begingroup$
                                                      1 byte shorter with set.intersection.
                                                      $endgroup$
                                                      – ovs
                                                      19 hours ago
















                                                    $begingroup$
                                                    1 byte shorter with set.intersection.
                                                    $endgroup$
                                                    – ovs
                                                    19 hours ago




                                                    $begingroup$
                                                    1 byte shorter with set.intersection.
                                                    $endgroup$
                                                    – ovs
                                                    19 hours ago











                                                    1












                                                    $begingroup$


                                                    Japt -h, 8 bytes



                                                    (I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)



                                                    mã rf iP


                                                    Try it or run all test cases



                                                    mã rf iP     :Implicit input of array
                                                    m :Map
                                                    ã : Substrings
                                                    r :Reduce by
                                                    f : Filter, keeping only elements that appear in both arrays
                                                    i :Prepend
                                                    P : An empty string
                                                    :Implicit output of last element





                                                    share|improve this answer











                                                    $endgroup$


















                                                      1












                                                      $begingroup$


                                                      Japt -h, 8 bytes



                                                      (I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)



                                                      mã rf iP


                                                      Try it or run all test cases



                                                      mã rf iP     :Implicit input of array
                                                      m :Map
                                                      ã : Substrings
                                                      r :Reduce by
                                                      f : Filter, keeping only elements that appear in both arrays
                                                      i :Prepend
                                                      P : An empty string
                                                      :Implicit output of last element





                                                      share|improve this answer











                                                      $endgroup$
















                                                        1












                                                        1








                                                        1





                                                        $begingroup$


                                                        Japt -h, 8 bytes



                                                        (I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)



                                                        mã rf iP


                                                        Try it or run all test cases



                                                        mã rf iP     :Implicit input of array
                                                        m :Map
                                                        ã : Substrings
                                                        r :Reduce by
                                                        f : Filter, keeping only elements that appear in both arrays
                                                        i :Prepend
                                                        P : An empty string
                                                        :Implicit output of last element





                                                        share|improve this answer











                                                        $endgroup$




                                                        Japt -h, 8 bytes



                                                        (I could knock off the last 3 bytes and use the -Fh flag instead but I'm not a fan of using -F)



                                                        mã rf iP


                                                        Try it or run all test cases



                                                        mã rf iP     :Implicit input of array
                                                        m :Map
                                                        ã : Substrings
                                                        r :Reduce by
                                                        f : Filter, keeping only elements that appear in both arrays
                                                        i :Prepend
                                                        P : An empty string
                                                        :Implicit output of last element






                                                        share|improve this answer














                                                        share|improve this answer



                                                        share|improve this answer








                                                        edited 19 hours ago

























                                                        answered 21 hours ago









                                                        ShaggyShaggy

                                                        18.9k21667




                                                        18.9k21667























                                                            1












                                                            $begingroup$

                                                            TSQL query, 154 bytes



                                                            USE master
                                                            DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)
                                                            INSERT @ values('string'),('stRIng');

                                                            SELECT top 1x FROM(SELECT
                                                            distinct substring(a,f.number,g.number)x,i
                                                            FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D
                                                            GROUP BY x ORDER BY-sum(i),-len(x)


                                                            Try it online






                                                            share|improve this answer











                                                            $endgroup$


















                                                              1












                                                              $begingroup$

                                                              TSQL query, 154 bytes



                                                              USE master
                                                              DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)
                                                              INSERT @ values('string'),('stRIng');

                                                              SELECT top 1x FROM(SELECT
                                                              distinct substring(a,f.number,g.number)x,i
                                                              FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D
                                                              GROUP BY x ORDER BY-sum(i),-len(x)


                                                              Try it online






                                                              share|improve this answer











                                                              $endgroup$
















                                                                1












                                                                1








                                                                1





                                                                $begingroup$

                                                                TSQL query, 154 bytes



                                                                USE master
                                                                DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)
                                                                INSERT @ values('string'),('stRIng');

                                                                SELECT top 1x FROM(SELECT
                                                                distinct substring(a,f.number,g.number)x,i
                                                                FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D
                                                                GROUP BY x ORDER BY-sum(i),-len(x)


                                                                Try it online






                                                                share|improve this answer











                                                                $endgroup$



                                                                TSQL query, 154 bytes



                                                                USE master
                                                                DECLARE @ table(a varchar(999)collate Latin1_General_CS_AI,i int identity)
                                                                INSERT @ values('string'),('stRIng');

                                                                SELECT top 1x FROM(SELECT
                                                                distinct substring(a,f.number,g.number)x,i
                                                                FROM spt_values f,spt_values g,@ WHERE'L'=g.type)D
                                                                GROUP BY x ORDER BY-sum(i),-len(x)


                                                                Try it online







                                                                share|improve this answer














                                                                share|improve this answer



                                                                share|improve this answer








                                                                edited 16 hours ago

























                                                                answered 19 hours ago









                                                                t-clausen.dkt-clausen.dk

                                                                1,994314




                                                                1,994314























                                                                    1












                                                                    $begingroup$

                                                                    Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes



                                                                    may be golfed more certainly



                                                                    map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h


                                                                    TIO






                                                                    share|improve this answer











                                                                    $endgroup$


















                                                                      1












                                                                      $begingroup$

                                                                      Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes



                                                                      may be golfed more certainly



                                                                      map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h


                                                                      TIO






                                                                      share|improve this answer











                                                                      $endgroup$
















                                                                        1












                                                                        1








                                                                        1





                                                                        $begingroup$

                                                                        Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes



                                                                        may be golfed more certainly



                                                                        map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h


                                                                        TIO






                                                                        share|improve this answer











                                                                        $endgroup$



                                                                        Perl 5 (-aln0777F/n/ -M5.01 -MList::util=max), 99 bytes



                                                                        may be golfed more certainly



                                                                        map/(.+)(?!.*1)(?{$h{$&}++})(?!)/,@F;say for grep{y///c==max map y///c,@b}@b=grep@F==$h{$_},keys%h


                                                                        TIO







                                                                        share|improve this answer














                                                                        share|improve this answer



                                                                        share|improve this answer








                                                                        edited 14 hours ago

























                                                                        answered 14 hours ago









                                                                        Nahuel FouilleulNahuel Fouilleul

                                                                        2,925211




                                                                        2,925211























                                                                            0












                                                                            $begingroup$


                                                                            Java (JDK), 176 bytes





                                                                            a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}


                                                                            Try it online!



                                                                            This is a rather naive implementation.






                                                                            share|improve this answer









                                                                            $endgroup$


















                                                                              0












                                                                              $begingroup$


                                                                              Java (JDK), 176 bytes





                                                                              a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}


                                                                              Try it online!



                                                                              This is a rather naive implementation.






                                                                              share|improve this answer









                                                                              $endgroup$
















                                                                                0












                                                                                0








                                                                                0





                                                                                $begingroup$


                                                                                Java (JDK), 176 bytes





                                                                                a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}


                                                                                Try it online!



                                                                                This is a rather naive implementation.






                                                                                share|improve this answer









                                                                                $endgroup$




                                                                                Java (JDK), 176 bytes





                                                                                a->{int l=a.get(0).length(),m=0,i=0,j;var r="";for(;i<l;i++)for(j=i;j++<l;){var s=a.get(0).substring(i,j);if(j-i>m&a.stream().allMatch(x->x.contains(s))){r=s;m=j-i;}}return r;}


                                                                                Try it online!



                                                                                This is a rather naive implementation.







                                                                                share|improve this answer












                                                                                share|improve this answer



                                                                                share|improve this answer










                                                                                answered 21 hours ago









                                                                                Olivier GrégoireOlivier Grégoire

                                                                                9,31511944




                                                                                9,31511944























                                                                                    0












                                                                                    $begingroup$


                                                                                    C# (Visual C# Interactive Compiler), 147 145 bytes





                                                                                    a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}


                                                                                    Try it online!






                                                                                    share|improve this answer











                                                                                    $endgroup$


















                                                                                      0












                                                                                      $begingroup$


                                                                                      C# (Visual C# Interactive Compiler), 147 145 bytes





                                                                                      a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}


                                                                                      Try it online!






                                                                                      share|improve this answer











                                                                                      $endgroup$
















                                                                                        0












                                                                                        0








                                                                                        0





                                                                                        $begingroup$


                                                                                        C# (Visual C# Interactive Compiler), 147 145 bytes





                                                                                        a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}


                                                                                        Try it online!






                                                                                        share|improve this answer











                                                                                        $endgroup$




                                                                                        C# (Visual C# Interactive Compiler), 147 145 bytes





                                                                                        a=>{int i=0,j,m=0,k=a[0].Length;string s="",d=s;for(;i<k;i++)for(j=m;j++<k-i;)if(a.All(y=>y.Contains(s=a[0].Substring(i,j)))){m=j;d=s;}return d;}


                                                                                        Try it online!







                                                                                        share|improve this answer














                                                                                        share|improve this answer



                                                                                        share|improve this answer








                                                                                        edited 15 hours ago

























                                                                                        answered 15 hours ago









                                                                                        Expired DataExpired Data

                                                                                        2686




                                                                                        2686























                                                                                            0












                                                                                            $begingroup$

                                                                                            JavaScript (ES6),  98  92 bytes





                                                                                            a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])


                                                                                            Try it online!






                                                                                            share|improve this answer











                                                                                            $endgroup$


















                                                                                              0












                                                                                              $begingroup$

                                                                                              JavaScript (ES6),  98  92 bytes





                                                                                              a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])


                                                                                              Try it online!






                                                                                              share|improve this answer











                                                                                              $endgroup$
















                                                                                                0












                                                                                                0








                                                                                                0





                                                                                                $begingroup$

                                                                                                JavaScript (ES6),  98  92 bytes





                                                                                                a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])


                                                                                                Try it online!






                                                                                                share|improve this answer











                                                                                                $endgroup$



                                                                                                JavaScript (ES6),  98  92 bytes





                                                                                                a=>(g=b=s=>a.every(x=>~x.indexOf(s))?b=b[s.length]?b:s:g(s.slice(0,-1,g(s.slice(1)))))(a[0])


                                                                                                Try it online!







                                                                                                share|improve this answer














                                                                                                share|improve this answer



                                                                                                share|improve this answer








                                                                                                edited 13 hours ago

























                                                                                                answered 15 hours ago









                                                                                                ArnauldArnauld

                                                                                                79.6k797330




                                                                                                79.6k797330























                                                                                                    0












                                                                                                    $begingroup$


                                                                                                    Red, 266 174 bytes



                                                                                                    func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u
                                                                                                    next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]


                                                                                                    Try it online!



                                                                                                    Changed the recursion to iteration and got rid of the sorting.






                                                                                                    share|improve this answer











                                                                                                    $endgroup$


















                                                                                                      0












                                                                                                      $begingroup$


                                                                                                      Red, 266 174 bytes



                                                                                                      func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u
                                                                                                      next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]


                                                                                                      Try it online!



                                                                                                      Changed the recursion to iteration and got rid of the sorting.






                                                                                                      share|improve this answer











                                                                                                      $endgroup$
















                                                                                                        0












                                                                                                        0








                                                                                                        0





                                                                                                        $begingroup$


                                                                                                        Red, 266 174 bytes



                                                                                                        func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u
                                                                                                        next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]


                                                                                                        Try it online!



                                                                                                        Changed the recursion to iteration and got rid of the sorting.






                                                                                                        share|improve this answer











                                                                                                        $endgroup$




                                                                                                        Red, 266 174 bytes



                                                                                                        func[b][l: length? s: b/1 n: 1 until[i: 1 loop n[t: copy/part at s i l r: on foreach u
                                                                                                        next b[r: r and(none <> find/case u t)]if r[return t]i: i + 1]n: n + 1 1 > l: l - 1]""]


                                                                                                        Try it online!



                                                                                                        Changed the recursion to iteration and got rid of the sorting.







                                                                                                        share|improve this answer














                                                                                                        share|improve this answer



                                                                                                        share|improve this answer








                                                                                                        edited 11 hours ago

























                                                                                                        answered 18 hours ago









                                                                                                        Galen IvanovGalen Ivanov

                                                                                                        7,26211034




                                                                                                        7,26211034























                                                                                                            0












                                                                                                            $begingroup$


                                                                                                            Perl 5, 87 bytes





                                                                                                            my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r


                                                                                                            Try it online!






                                                                                                            share|improve this answer









                                                                                                            $endgroup$


















                                                                                                              0












                                                                                                              $begingroup$


                                                                                                              Perl 5, 87 bytes





                                                                                                              my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r


                                                                                                              Try it online!






                                                                                                              share|improve this answer









                                                                                                              $endgroup$
















                                                                                                                0












                                                                                                                0








                                                                                                                0





                                                                                                                $begingroup$


                                                                                                                Perl 5, 87 bytes





                                                                                                                my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r


                                                                                                                Try it online!






                                                                                                                share|improve this answer









                                                                                                                $endgroup$




                                                                                                                Perl 5, 87 bytes





                                                                                                                my$r;"$&@_"=~/(.{@{[$r=~y,,,c]},}).*(n.*1.*){@{[@_-1]}}/ and$r=$1while$_[0]=~s,.,,;$r


                                                                                                                Try it online!







                                                                                                                share|improve this answer












                                                                                                                share|improve this answer



                                                                                                                share|improve this answer










                                                                                                                answered 8 hours ago









                                                                                                                Kjetil S.Kjetil S.

                                                                                                                58925




                                                                                                                58925























                                                                                                                    0












                                                                                                                    $begingroup$


                                                                                                                    Retina 0.8.2, 48 bytes



                                                                                                                    M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)
                                                                                                                    O#$^`
                                                                                                                    $.&
                                                                                                                    1G`


                                                                                                                    Try it online! Explanation:



                                                                                                                    M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)


                                                                                                                    For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.



                                                                                                                    O#$^`
                                                                                                                    $.&


                                                                                                                    Sort the substrings in reverse order of length.



                                                                                                                    1G`


                                                                                                                    Keep only the first, i.e. the longest substring.






                                                                                                                    share|improve this answer











                                                                                                                    $endgroup$


















                                                                                                                      0












                                                                                                                      $begingroup$


                                                                                                                      Retina 0.8.2, 48 bytes



                                                                                                                      M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)
                                                                                                                      O#$^`
                                                                                                                      $.&
                                                                                                                      1G`


                                                                                                                      Try it online! Explanation:



                                                                                                                      M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)


                                                                                                                      For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.



                                                                                                                      O#$^`
                                                                                                                      $.&


                                                                                                                      Sort the substrings in reverse order of length.



                                                                                                                      1G`


                                                                                                                      Keep only the first, i.e. the longest substring.






                                                                                                                      share|improve this answer











                                                                                                                      $endgroup$
















                                                                                                                        0












                                                                                                                        0








                                                                                                                        0





                                                                                                                        $begingroup$


                                                                                                                        Retina 0.8.2, 48 bytes



                                                                                                                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)
                                                                                                                        O#$^`
                                                                                                                        $.&
                                                                                                                        1G`


                                                                                                                        Try it online! Explanation:



                                                                                                                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)


                                                                                                                        For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.



                                                                                                                        O#$^`
                                                                                                                        $.&


                                                                                                                        Sort the substrings in reverse order of length.



                                                                                                                        1G`


                                                                                                                        Keep only the first, i.e. the longest substring.






                                                                                                                        share|improve this answer











                                                                                                                        $endgroup$




                                                                                                                        Retina 0.8.2, 48 bytes



                                                                                                                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)
                                                                                                                        O#$^`
                                                                                                                        $.&
                                                                                                                        1G`


                                                                                                                        Try it online! Explanation:



                                                                                                                        M&!`(?<=^.*)(.+)(?=(.*n.*1)*.*$)


                                                                                                                        For each suffix of the first string, find the longest prefix that's also a substring of all of the other strings. List all of those suffix prefixes (i.e. substrings). If there are no matching substrings, we just end up with the empty string, which is what we want anyway.



                                                                                                                        O#$^`
                                                                                                                        $.&


                                                                                                                        Sort the substrings in reverse order of length.



                                                                                                                        1G`


                                                                                                                        Keep only the first, i.e. the longest substring.







                                                                                                                        share|improve this answer














                                                                                                                        share|improve this answer



                                                                                                                        share|improve this answer








                                                                                                                        edited 6 hours ago

























                                                                                                                        answered 6 hours ago









                                                                                                                        NeilNeil

                                                                                                                        82k745178




                                                                                                                        82k745178























                                                                                                                            0












                                                                                                                            $begingroup$


                                                                                                                            JavaScript (Node.js), 106 bytes





                                                                                                                            a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.indexOf(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)


                                                                                                                            Try it online!





                                                                                                                            a=>(                      // Main function
                                                                                                                            F=( // Helper function to run through all substrings in a[0]
                                                                                                                            l, // Length
                                                                                                                            n, // Start position
                                                                                                                            w=a[0].substr(n,l) // The substring
                                                                                                                            )=>
                                                                                                                            l? // If l > 0:
                                                                                                                            n<0? // If n < 0:
                                                                                                                            F(--l,L-l) // Check another length
                                                                                                                            :a.some( // If n >= 0:
                                                                                                                            y=>y.indexOf(w)<0 // Check whether there is any string not containing the substring
                                                                                                                            // (indexOf used because of presence of regex special characters)
                                                                                                                            )? // If so:
                                                                                                                            F(l,n-1) // Check another substring
                                                                                                                            :w // If not, return this substring and terminate
                                                                                                                            // (This function checks from the longest substring possible, so
                                                                                                                            // it is safe to return right here)
                                                                                                                            :"" // If l <= 0: Return empty string (no common substring)
                                                                                                                            )(
                                                                                                                            L=a[0].length, // Starts from length = the whole length of a[0]
                                                                                                                            0 // And start position = 0
                                                                                                                            )





                                                                                                                            share|improve this answer











                                                                                                                            $endgroup$


















                                                                                                                              0












                                                                                                                              $begingroup$


                                                                                                                              JavaScript (Node.js), 106 bytes





                                                                                                                              a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.indexOf(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)


                                                                                                                              Try it online!





                                                                                                                              a=>(                      // Main function
                                                                                                                              F=( // Helper function to run through all substrings in a[0]
                                                                                                                              l, // Length
                                                                                                                              n, // Start position
                                                                                                                              w=a[0].substr(n,l) // The substring
                                                                                                                              )=>
                                                                                                                              l? // If l > 0:
                                                                                                                              n<0? // If n < 0:
                                                                                                                              F(--l,L-l) // Check another length
                                                                                                                              :a.some( // If n >= 0:
                                                                                                                              y=>y.indexOf(w)<0 // Check whether there is any string not containing the substring
                                                                                                                              // (indexOf used because of presence of regex special characters)
                                                                                                                              )? // If so:
                                                                                                                              F(l,n-1) // Check another substring
                                                                                                                              :w // If not, return this substring and terminate
                                                                                                                              // (This function checks from the longest substring possible, so
                                                                                                                              // it is safe to return right here)
                                                                                                                              :"" // If l <= 0: Return empty string (no common substring)
                                                                                                                              )(
                                                                                                                              L=a[0].length, // Starts from length = the whole length of a[0]
                                                                                                                              0 // And start position = 0
                                                                                                                              )





                                                                                                                              share|improve this answer











                                                                                                                              $endgroup$
















                                                                                                                                0












                                                                                                                                0








                                                                                                                                0





                                                                                                                                $begingroup$


                                                                                                                                JavaScript (Node.js), 106 bytes





                                                                                                                                a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.indexOf(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)


                                                                                                                                Try it online!





                                                                                                                                a=>(                      // Main function
                                                                                                                                F=( // Helper function to run through all substrings in a[0]
                                                                                                                                l, // Length
                                                                                                                                n, // Start position
                                                                                                                                w=a[0].substr(n,l) // The substring
                                                                                                                                )=>
                                                                                                                                l? // If l > 0:
                                                                                                                                n<0? // If n < 0:
                                                                                                                                F(--l,L-l) // Check another length
                                                                                                                                :a.some( // If n >= 0:
                                                                                                                                y=>y.indexOf(w)<0 // Check whether there is any string not containing the substring
                                                                                                                                // (indexOf used because of presence of regex special characters)
                                                                                                                                )? // If so:
                                                                                                                                F(l,n-1) // Check another substring
                                                                                                                                :w // If not, return this substring and terminate
                                                                                                                                // (This function checks from the longest substring possible, so
                                                                                                                                // it is safe to return right here)
                                                                                                                                :"" // If l <= 0: Return empty string (no common substring)
                                                                                                                                )(
                                                                                                                                L=a[0].length, // Starts from length = the whole length of a[0]
                                                                                                                                0 // And start position = 0
                                                                                                                                )





                                                                                                                                share|improve this answer











                                                                                                                                $endgroup$




                                                                                                                                JavaScript (Node.js), 106 bytes





                                                                                                                                a=>(F=(l,n,w=a[0].substr(n,l))=>l?n<0?F(--l,L-l):a.some(y=>y.indexOf(w)<0)?F(l,n-1):w:"")(L=a[0].length,0)


                                                                                                                                Try it online!





                                                                                                                                a=>(                      // Main function
                                                                                                                                F=( // Helper function to run through all substrings in a[0]
                                                                                                                                l, // Length
                                                                                                                                n, // Start position
                                                                                                                                w=a[0].substr(n,l) // The substring
                                                                                                                                )=>
                                                                                                                                l? // If l > 0:
                                                                                                                                n<0? // If n < 0:
                                                                                                                                F(--l,L-l) // Check another length
                                                                                                                                :a.some( // If n >= 0:
                                                                                                                                y=>y.indexOf(w)<0 // Check whether there is any string not containing the substring
                                                                                                                                // (indexOf used because of presence of regex special characters)
                                                                                                                                )? // If so:
                                                                                                                                F(l,n-1) // Check another substring
                                                                                                                                :w // If not, return this substring and terminate
                                                                                                                                // (This function checks from the longest substring possible, so
                                                                                                                                // it is safe to return right here)
                                                                                                                                :"" // If l <= 0: Return empty string (no common substring)
                                                                                                                                )(
                                                                                                                                L=a[0].length, // Starts from length = the whole length of a[0]
                                                                                                                                0 // And start position = 0
                                                                                                                                )






                                                                                                                                share|improve this answer














                                                                                                                                share|improve this answer



                                                                                                                                share|improve this answer








                                                                                                                                edited 5 hours ago

























                                                                                                                                answered yesterday









                                                                                                                                Shieru AsakotoShieru Asakoto

                                                                                                                                2,750317




                                                                                                                                2,750317






























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