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Longest common substring

From Rosetta Code
Task
Longest common substring
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Write a function that returns the longest common substring of two strings.

Use it within a program that demonstrates sample output from the function, which will consist of the longest common substring between "thisisatest" and "testing123testing".

Note that substrings are consecutive characters within a string.   This distinguishes them from subsequences, which is any sequence of characters within a string, even if there are extraneous characters in between them.

Hence, the longest common subsequence between "thisisatest" and "testing123testing" is "tsitest", whereas the longest common substring is just "test".


Other tasks related to string operations:
Metrics
Counting
Remove/replace
Anagrams/Derangements/shuffling
Find/Search/Determine
Formatting
Song lyrics/poems/Mad Libs/phrases
Tokenize
Sequences


References



11l[edit]

Translation of: Python
F longest_common_substring(s1, s2)
   V ir = 0
   V jr = -1
   L(i1) 0 .< s1.len
      V? i2 = s2.find(s1[i1])
      L i2 != N
         V (j1, j2) = (i1, i2)
         L j1 < s1.len & j2 < s2.len & s2[j2] == s1[j1]
            I j1 - i1 >= jr - ir
               (ir, jr) = (i1, j1)
            j1++
            j2++
         i2 = s2.find(s1[i1], i2 + 1)
   R s1[ir..jr]
 
print(longest_common_substring(‘thisisatest’, ‘testing123testing’))
Output:
test

Ada[edit]

with Ada.Text_Io;
 
procedure Longest_Common_Substring is
 
   function Common (Left, Right: String) return String is
      Com : array (Left'Range, Right'Range) of Natural := (others => (others => 0));
      Longest : Natural := 0;
      Last    : Natural := 0;
   begin
      for L in Left'Range loop
         for R in Right'Range loop
            if Left (L) = Right (R) then
               if L > Left'First and R > Right'First then
                  Com (L, R) := Com (L - 1, R - 1) + 1;
               else
                  Com (L, R) := 1;
               end if;
               if Com (L, R) > Longest then
                  Longest := Com (L, R);
                  Last    := L;
               end if;
            end if;
         end loop;
      end loop;
      return Left (Last - Longest + 1 .. Last);
   end Common;
 
begin
   Ada.Text_Io.Put_Line (Common ("thisisatest", "testing123testing"));
end Longest_Common_Substring;
Output:
test

Aime[edit]

void
test_string(text &g, v, l)
{
    integer n;
 
    n = prefix(v, l);
    if (~g < n) {
        g = cut(l, 0, n);
    }
}
 
longest(text u, v)
{
    record r;
    text g, l, s;
 
    while (~u) {
        r[u] = 0;
        u = delete(u, 0);
    }
    while (~v) {
        if (rsk_lower(r, v, l)) {
            test_string(g, v, l);
        }
        if (rsk_upper(r, v, l)) {
            test_string(g, v, l);
        }
        v = delete(v, 0);
    }
 
    g;
}
o_(longest("thisisatest", "testing123testing"), "\n");

ALGOL 68[edit]

BEGIN
    # returns the longest common substring of s and t #
    PROC longest common substring = ( STRING s, t )STRING:
         BEGIN
            STRING s1       = s[ @ 1 ]; # normalise bounds to 1 : ... #
            STRING s2       = t[ @ 1 ];
            STRING result  := "";
            INT result len := 0;
            FOR i TO UPB s1 DO
                FOR j TO UPB s2 DO
                    IF s1[ i ] = s2[ j ] THEN
                        INT k := 1;
                        WHILE INT ik = i + k;
                              INT jk = j + k;
                              IF ik > UPB s1 OR jk > UPB s2
                              THEN FALSE
                              ELSE s1[ ik ] = s2[ jk ]
                              FI
                        DO
                            k +:= 1
                        OD;
                        IF k > result len THEN
                            # found a longer substring #
                            result len := k;
                            result     := s1[ i : ( i + k ) - 1 ]
                        FI
                    FI
                OD
            OD;
            result
         END # longest common substring # ;
 
    # task test case #
    print( ( longest common substring( "thisisatest", "testing123testing" ), newline ) )
END
Output:
test

AppleScript[edit]

Iterative[edit]

This allows for the possibility of co-longest substrings, returning one instance of each. If either input string is empty, it's taken as meaning there are no common substrings. 

on LCS(a, b)
    -- Identify the shorter string. The longest common substring won't be longer than it!
    set lengthA to a's length
    set lengthB to b's length
    if (lengthA < lengthB) then
        set {shorterString, shorterLength, longerString} to {a, lengthA, b}
    else
        set {shorterString, shorterLength, longerString} to {b, lengthB, a}
    end if
 
    set longestMatches to {}
    set longestMatchLength to 0
    -- Find the longest matching substring starting at each character in the shorter string.
    repeat with i from 1 to shorterLength
        repeat with j from shorterLength to i by -1
            set thisSubstring to text i thru j of shorterString
            if (longerString contains thisSubstring) then
                -- Match found. If it's longer than the previously found match, or a new string of the same length, remember it.
                set matchLength to j - i + 1
                if (matchLength > longestMatchLength) then
                    set longestMatches to {thisSubstring}
                    set longestMatchLength to matchLength
                else if ((matchLength = longestMatchLength) and (thisSubstring is not in longestMatches)) then
                    set end of longestMatches to thisSubstring
                end if
                -- Don't bother with the match's own substrings.
                exit repeat
            end if
        end repeat
    end repeat
 
    return longestMatches
end LCS
LCS("thisisatest", "testing123testing")
Output:
{"test"}

Or: 

LCS("thisisthebesttest", "besting123testing")
Output:
{"best", "test"}

Functional[edit]

Using library functions wherever possible, for better productivity, (and for more granular Rosetta comparison): 

------------------ LONGEST COMMON SUBSTRING ----------------
 
-- longestCommon :: Eq a => [a] -> [a] -> [a]
on longestCommon(a, b)
    -- The longest common substring of two given strings.
    script substrings
        on |λ|(s)
            map(my concat, concatMap(my tails, rest of inits(s)))
        end |λ|
    end script
 
    set {xs, ys} to map(substrings, {a, b})
    maximumBy(comparing(my |length|), intersect(xs, ys))
end longestCommon
 
 
 
-------------------------- TEST ---------------------------
on run
    longestCommon("testing123testing", "thisisatest")
end run
 
 
 
-------------------- GENERIC FUNCTIONS --------------------
 
-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
    script
        on |λ|(a, b)
            tell mReturn(f)
                set fa to |λ|(a)
                set fb to |λ|(b)
                if fa < fb then
                    -1
                else if fa > fb then
                    1
                else
                    0
                end if
            end tell
        end |λ|
    end script
end comparing
 
 
-- concat :: [String] -> String
on concat(xs)
    script go
        on |λ|(a, x)
            a & x
        end |λ|
    end script
    foldl(go, "", xs)
end concat
 
 
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
    set lng to length of xs
    set acc to {}
    tell mReturn(f)
        repeat with i from 1 to lng
            set acc to acc & (|λ|(item i of xs, i, xs))
        end repeat
    end tell
    return acc
end concatMap
 
 
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
    tell mReturn(f)
        set v to startValue
        set lng to length of xs
        repeat with i from 1 to lng
            set v to |λ|(v, item i of xs, i, xs)
        end repeat
        return v
    end tell
end foldl
 
 
-- inits :: String -> [String]
on inits(xs)
    script charInit
        on |λ|(_, i, xs)
            text 1 thru i of xs
        end |λ|
    end script
 
    {""} & map(charInit, xs)
end inits
 
 
-- intersect :: (Eq a) => [a] -> [a] -> [a]
on intersect(xs, ys)
    if length of xs < length of ys then
        set {shorter, longer} to {xs, ys}
    else
        set {longer, shorter} to {xs, ys}
    end if
    if shorter ≠ {} then
        set lst to {}
        repeat with x in shorter
            if longer contains x then set end of lst to contents of x
        end repeat
        lst
    else
        {}
    end if
end intersect
 
 
-- length :: [a] -> Int
on |length|(xs)
    set c to class of xs
    if list is c or string is c then
        length of xs
    else
        (2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
    end if
end |length|
 
 
-- maximumBy :: (a -> a -> Ordering) -> [a] -> a
on maximumBy(f, xs)
    set cmp to mReturn(f)
    script max
        on |λ|(a, b)
            if a is missing value or cmp's |λ|(a, b) < 0 then
                b
            else
                a
            end if
        end |λ|
    end script
 
    foldl(max, missing value, xs)
end maximumBy
 
 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
    -- 2nd class handler function lifted into 1st class script wrapper. 
    if script is class of f then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    -- The list obtained by applying f
    -- to each element of xs.
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map
 
 
-- tails :: String -> [String]
on tails(xs)
    set es to characters of xs
    script residue
        on |λ|(_, i)
            items i thru -1 of es
        end |λ|
    end script
    map(residue, es) & {""}
end tails
Output:
"test"

Arturo[edit]

lcs: function [a,b][
    lengths: map 0..size a => [map 0..size b => 0]
    greatestLength: 0
    result: ""
    loop.with:'i a 'x [
        loop.with:'j b 'y [
            if x=y [
                if? or? i=0 j=0 ->
                    lengths\[i]\[j]: 0
                else ->
                    lengths\[i]\[j]: 1 + lengths\[i-1]\[j-1]
 
                if greatestLength < lengths\[i]\[j] [
                    greatestLength: lengths\[i]\[j]
                    result: slice a (i-greatestLength)+1 i
                ]
            ]
        ]
    ]
    return result
]
 
print lcs "thisisatest", "testing123testing"
Output:
test

AutoHotkey[edit]

Using Text Comparison[edit]

LCS(a, b){
	x := i := 1
	while StrLen(x)	
		Loop % StrLen(a)
			IfInString, b, % x := SubStr(a, i:=StrLen(x)=1 ? i+1 : i, n:=StrLen(a)+1-A_Index)
				res := StrLen(res) > StrLen(x) ? res : x 
	return res
}
Examples:
MsgBox % LCS("thisisatest", "testing123testing")
Outputs:
test

Using RegEx[edit]

LCS(a, b){
	while pos := RegExMatch(a "`n" b, "(.+)(?=.*\R.*\1)", m, pos?pos+StrLen(m):1)
		res := StrLen(res) > StrLen(m1) ? res : m1
	return res
}
Examples:
MsgBox % LCS("thisisatest", "testing123testing")
Outputs:
test

BaCon[edit]

FUNCTION Common_Sub$(haystack$, needle$)
 
    WHILE LEN(needle$)
        FOR x = LEN(needle$) DOWNTO 1
            IF INSTR(haystack$, LEFT$(needle$, x)) THEN RETURN LEFT$(needle$, x)
        NEXT
        needle$ = MID$(needle$, 2)
    WEND
    EXIT
 
ENDFUNC
 
PRINT Common_Sub$("thisisatest", "testing123testing")
Output:
test


BASIC256[edit]

Translation of: FreeBASIC
function LCS(a, b)
	if length(a) = 0 or length(b) = 0 then return ""
	while length(b)
		for j = length(b) to 1 step -1
			if instr(a, left(b, j)) then return left(b, j)
		next j
		b = mid$(b, 2)
	end while
end function
 
print LCS("thisisatest", "testing123testing")
end


Bracmat[edit]

  ( lcs
  =   X a b x L
    .   !arg:(?a.?b)
      & 0:?L
      & :?X
      & ( @( !a
           :   ?
               ( ?x
               & @(!x:? [>!L)
               & @(!b:? !x ?)
               & @(!x:? [?L:?X)
               )
               (?&~)
           )
        | !X
        )
  )
& out$(lcs$(thisisatest.testing123testing))

Output 

test

C[edit]

Translation of: Modula-2
#include <stdio.h>
 
void lcs(const char * const sa, const char * const sb, char ** const beg, char ** const end) {
    size_t apos, bpos;
    ptrdiff_t len;
 
    *beg = 0;
    *end = 0;
    len = 0;
 
    for (apos = 0; sa[apos] != 0; ++apos) {
        for (bpos = 0; sb[bpos] != 0; ++bpos) {
            if (sa[apos] == sb[bpos]) {
                len = 1;
                while (sa[apos + len] != 0 && sb[bpos + len] != 0 && sa[apos + len] == sb[bpos + len]) {
                    len++;
                }
            }
 
            if (len > *end - *beg) {
                *beg = sa + apos;
                *end = *beg + len;
                len = 0;
            }
        }
    }
}
 
int main() {
    char *s1 = "thisisatest";
    char *s2 = "testing123testing";
    char *beg, *end, *it;
 
    lcs(s1, s2, &beg, &end);
 
    for (it = beg; it != end; it++) {
        putchar(*it);
    }
    printf("\n");
 
    return 0;
}
Output:
test

C#[edit]

Using dynamic programming[edit]

using System;
 
namespace LongestCommonSubstring
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine(lcs("thisisatest", "testing123testing"));
            Console.ReadKey(true);
        }
 
        public static string lcs(string a, string b)
        {
            var lengths = new int[a.Length, b.Length];
            int greatestLength = 0;
            string output = "";
            for (int i = 0; i < a.Length; i++)
            {
                for (int j = 0; j < b.Length; j++)
                {
                    if (a[i] == b[j])
                    {
                        lengths[i, j] = i == 0 || j == 0 ? 1 : lengths[i - 1, j - 1] + 1;
                        if (lengths[i, j] > greatestLength)
                        {
                            greatestLength = lengths[i, j];
                            output = a.Substring(i - greatestLength + 1, greatestLength);
                        }
                    }
                    else
                    {
                        lengths[i, j] = 0;
                    }
                }
            }
            return output;
        }
    }
}
Output:
test

Searching for smaller substrings of a in b[edit]

Translation of: REXX
//C# program tests the LCSUBSTR (Longest Common Substring) subroutine.
using System;
namespace LongestCommonSubstring
{
    class Program
    {
        static void Main(string[] args)
        {
            string a = args.Length >= 1 ? args[0] : "";                                             /*get two arguments (strings).   */
            string b = args.Length == 2 ? args[1] : "";
            if (a == "") a = "thisisatest";                                                         /*use this string for a default. */
            if (b == "") b = "testing123testing";                                                   /* "    "     "    "  "    "     */
            Console.WriteLine("string A = {0}", a);                                                 /*echo string  A  to screen.     */
            Console.WriteLine("string B = {0}", b);                                                 /*echo string  B  to screen.     */
            Console.WriteLine("LCsubstr = {0}", LCsubstr(a, b));                                    /*tell Longest Common Substring. */
            Console.ReadKey(true);
        }                                                                                           /*stick a fork in it, we're done.*/
 
        /*─────────────────────────────────LCSUBSTR subroutine─────────────────────────────────*/
        public static string LCsubstr(string x, string y)                                           /*Longest Common Substring.      */
        {
            string output = "";
            int lenx = x.Length;                                                                    /*shortcut for using the X length*/
            for (int j = 0; j < lenx; j++)                                                          /*step through start points in X.*/
            {
                for (int k = lenx - j; k > -1; k--)                                                 /*step through string lengths.   */
                {
                    string common = x.Substring(j, k);                                              /*extract a common substring.    */
                    if (y.IndexOf(common) > -1 && common.Length > output.Length) output = common;   /*longest?*/
                }                                                                                   /*k*/
            }                                                                                       /*j*/
            return output;                                                                          /*$  is "" if no common string.  */
        }
    }
}

output when using the default inputs: 

   string A = thisisatest
   string B = testing123testing
   LCsubstr = test

Searching for smaller substrings of a in b (simplified)[edit]

Translation of: zkl
//C# program tests the LCS (Longest Common Substring) subroutine.
using System;
namespace LongestCommonSubstring
{
    class Program
    {
        static void Main(string[] args)
        {
            string a = args.Length >= 1 ? args[0] : "";                                             /*get two arguments (strings).   */
            string b = args.Length == 2 ? args[1] : "";
            if (a == "") a = "thisisatest";                                                         /*use this string for a default. */
            if (b == "") b = "testing123testing";                                                   /* "    "     "    "  "    "     */
            Console.WriteLine("string A = {0}", a);                                                 /*echo string  A  to screen.     */
            Console.WriteLine("string B = {0}", b);                                                 /*echo string  B  to screen.     */
            Console.WriteLine("LCS = {0}", lcs(a, b));                                              /*tell Longest Common Substring. */
            Console.ReadKey(true);
        }                                                                                           /*stick a fork in it, we're done.*/
 
        /*─────────────────────────────────LCS subroutine─────────────────────────────────*/
        private static string lcs(string a, string b)
        {
           if(b.Length<a.Length){ string t=a; a=b; b=t; }
           for (int n = a.Length; n > 0; n--)
           {
              for (int m = a.Length-n; m <= a.Length-n; m++)
              {
                  string s=a.Substring(m,n);
                  if(b.Contains(s)) return(s);
              }
           }
           return "";
        }    
    }
 

output when using the default inputs: 

   string A = thisisatest
   string B = testing123testing
   LCS = test

C++[edit]

Works with: C++14
#include <string>
#include <algorithm>
#include <iostream>
#include <set>
#include <vector>
 
auto collectSubStrings( const std::string& s, int maxSubLength )
{
    int l = s.length();
    auto res = std::set<std::string>();
    for ( int start = 0; start < l; start++ )
    {
        int m = std::min( maxSubLength, l - start + 1 );
        for ( int length = 1; length < m; length++ )
        {
            res.insert( s.substr( start, length ) );
        }
    }
    return res;
}
 
std::string lcs( const std::string& s0, const std::string& s1 )
{
    // collect substring set
    auto maxSubLength = std::min( s0.length(), s1.length() );
    auto set0 = collectSubStrings( s0, maxSubLength );
    auto set1 = collectSubStrings( s1, maxSubLength );
 
    // get commons into a vector
    auto common = std::vector<std::string>();
    std::set_intersection( set0.begin(), set0.end(), set1.begin(), set1.end(),
        std::back_inserter( common ) );
 
    // get the longest one
    std::nth_element( common.begin(), common.begin(), common.end(),
        []( const std::string& s1, const std::string& s2 ) {
            return s1.length() > s2.length();
        } );
    return *common.begin();
}
 
int main( int argc, char* argv[] )
{
    auto s1 = std::string( "thisisatest" );
    auto s2 = std::string( "testing123testing" );
    std::cout << "The longest common substring of " << s1 << " and " << s2
              << " is:\n";
    std::cout << "\"" << lcs( s1, s2 ) << "\" !\n";
    return 0;
}
Output:
The longest common substring of thisisatest and testing123testing is:
"test" !

Common Lisp[edit]

 
(defun longest-common-substring (a b)
 "Return the longest substring common to a and b"
 ;; Found at https://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Longest_common_substring#Common_Lisp
  (let ((L (make-array (list (length a) (length b)) :initial-element 0))
        (z 0)
        (result '()) )
    (dotimes (i (length a))
      (dotimes (j (length b))
        (when (char= (char a i) (char b j))
          (setf (aref L i j)
                (if (or (zerop i) (zerop j))
                    1
                    (1+ (aref L (1- i) (1- j))) ))
          (when (> (aref L i j) z)
            (setf z (aref L i j)
                  result '() ))
          (when (= (aref L i j) z)
            (pushnew (subseq a (1+ (- i z)) (1+ i))
                     result :test #'equal )))))
    result ))
 
Output:
(longest-common-substring "thisisatest" "testing123testing") => ("test")


D[edit]

Translation of: C#
import std.stdio;
 
string lcs(string a, string b) {
    int[][] lengths;
    lengths.length = a.length;
    for (int i=0; i<a.length; i++) {
        lengths[i].length = b.length;
    }
 
    int greatestLength;
    string output;
    for (int i=0; i<a.length; i++) {
        for (int j=0; j<b.length; j++) {
            if (a[i]==b[j]) {
                lengths[i][j] = i==0 || j==0 ? 1 : lengths[i-1][j-1] + 1;
                if (lengths[i][j] > greatestLength) {
                    greatestLength = lengths[i][j];
                    int start = i-greatestLength+1;
                    output = a[start..start+greatestLength];
                }
            } else {
                lengths[i][j] = 0;
            }
        }
    }
    return output;
}
 
void main() {
    writeln(lcs("testing123testing", "thisisatest"));
}
Output:
test

Delphi[edit]

Translation of: C#
 
program Longest_Common_Substring;
 
{$APPTYPE CONSOLE}
 
{$R *.res}
 
uses
  System.SysUtils;
 
function lcs(x, y: string): string;
var
  n, m, Alength: Integer;
  t, common: string;
  j: Integer;
  k: Integer;
begin
 
  Result := '';
  Alength := x.Length;
 
  for j := 0 to Alength - 1 do
    for k := Alength - j downto 0 do
    begin
      common := x.Substring(j, k);
      if (y.IndexOf(common) > -1) and (common.Length > Result.Length) then
        Result := common;
    end;
end;
 
var
  a, b: string;
 
begin
  a := 'thisisatest';
  b := 'testing123testing';
  if ParamCount = 2 then
  begin
    if not ParamStr(1).IsEmpty then
      a := ParamStr(1);
    if not ParamStr(2).IsEmpty then
      b := ParamStr(2);
  end;
 
  Writeln('string A = ', a);
  Writeln('string B = ', b);
  Writeln('LCsubstr = ', lcs(a, b));
  readln;
end.
 
Output:
string A = thisisatest123
string B = testing123testing
LCsubstr = test

Dyalect[edit]

Translation of: Swift
func lComSubStr(w1, w2) {
    var (len, end) = (0, 0)
    var mat  = Array.empty(w1.len() + 1, () => Array.empty(w2.len() + 1, 0))
    var (i, j) = (0, 0)
 
    for sLett in w1 {
      for tLett in w2 {
        if tLett == sLett {
            let curLen = mat[i][j] + 1
            mat[i + 1][j + 1] = curLen
            if curLen > len {
            len = curLen
            end = i
            }
        }
        j += 1
      }
      j = 0
      i += 1
    }
    String(values: w1).substring((end + 1) - len, len)
}
 
func comSubStr(w1, w2) {
  return String(lComSubStr(w1.iter().toArray(), w2.iter().toArray()))
}
 
comSubStr("thisisatest", "testing123testing") // "test"

Elixir[edit]

Works with: Elixir version 1.3
defmodule LCS do
  def longest_common_substring(a,b) do
    alist = to_charlist(a) |> Enum.with_index
    blist = to_charlist(b) |> Enum.with_index
    lengths = for i <- 0..length(alist)-1, j <- 0..length(blist), into: %{}, do: {{i,j},0}
    Enum.reduce(alist, {lengths,0,""}, fn {x,i},acc ->
      Enum.reduce(blist, acc, fn {y,j},{map,gleng,lcs} ->
        if x==y do
          len = if i==0 or j==0, do: 1, else: map[{i-1,j-1}]+1
          map = %{map | {i,j} => len}
          if len > gleng, do: {map, len, String.slice(a, i - len + 1, len)},
                        else: {map, gleng, lcs}
        else
          {map, gleng, lcs}
        end
      end)
    end)
    |> elem(2)
  end
end
 
IO.puts LCS.longest_common_substring("thisisatest", "testing123testing")
Output:
test

Factor[edit]

Works with: Factor version 0.99 2020-07-03
USING: io sequences.extras ;
 
"thisisatest" "testing123testing" longest-subseq print
Output:
test


FreeBASIC[edit]

Function LCS(a As String, b As String) As String
    If Len(a) = 0 Or Len(b) = 0 Then Return ""
    While Len(b)
        For j As Integer = Len(b) To 1 Step -1
            If Instr(a, Left(b, j)) Then Return Left(b, j)
        Next j
        b = Mid(b, 2)
    Wend
End Function
 
Print LCS("thisisatest", "testing123testing")
Sleep


Go[edit]

Translation of: C#
package main
 
import "fmt"
 
func lcs(a, b string) (output string) {
    lengths := make([]int, len(a)*len(b))
    greatestLength := 0
    for i, x := range a {
        for j, y := range b {
            if x == y {
                if i == 0 || j == 0 {
                    lengths[i*len(b)+j] = 1
                } else {
                    lengths[i*len(b)+j] = lengths[(i-1)*len(b)+j-1] + 1
                }
                if lengths[i*len(b)+j] > greatestLength {
                    greatestLength = lengths[i*len(b)+j]
                    output = a[i-greatestLength+1 : i+1]
                }
            }
        }
    }
    return
}
 
func main() {
    fmt.Println(lcs("thisisatest", "testing123testing"))
}
Output:
test

Haskell[edit]

import Data.Ord (comparing)
import Data.List (maximumBy, intersect)
 
subStrings :: [a] -> [[a]]
subStrings s =
  let intChars = length s
  in [ take n $ drop i s
     | i <- [0 .. intChars - 1] 
     , n <- [1 .. intChars - i] ]
 
longestCommon :: Eq a => [a] -> [a] -> [a]
longestCommon a b =
  maximumBy (comparing length) (subStrings a `intersect` subStrings b)
 
main :: IO ()
main = putStrLn $ longestCommon "testing123testing" "thisisatest"
Output:
test

Or, fusing subStrings as tail . inits <=< tails 

import Control.Monad ((<=<))
import Data.List (inits, intersect, maximumBy, tails)
import Data.Ord (comparing)
 
----------------- LONGEST COMMON SUBSTRING ---------------
 
longestCommon :: Eq a => [a] -> [a] -> [a]
longestCommon a b =
  let pair [x, y] = (x, y)
   in maximumBy (comparing length) $
        (uncurry intersect . pair) $
          [tail . inits <=< tails] <*> [a, b]
 
--------------------------- TEST -------------------------
main :: IO ()
main =
  putStrLn $
    longestCommon "testing123testing" "thisisatest"
Output:
test

J[edit]

This algorithm starts by comparing each character in the one string to each character in the other, and then iterates on this result until it finds the length of the longest common substring. So if Lx is the length of one argument string, Ly is the length of the other argument string, and Lr is the length of the result string, this algorithm uses space on the order of Lx*Ly and time on the order of Lx*Ly*Lr.

In other words: this can be suitable for small problems, but you might want something better if you're comparing gigabyte length strings with high commonality. 

lcstr=:4 :0
  C=. ({.~ 1+$) x=/y
  M=. >./ (* * * >. * + (_1&|.)@:|:^:2)^:_  C
  N=. >./ M
  y {~ (M i. N)-i.-N
)

Intermedate results: 

   C shows which characters are in common between the two strings.
   M marks the length of the longest common substring ending at each position in the right argument
   N is the length of the longest common substring

Example use: 

   'thisisatest' lcs 'testing123testing'
test

Java[edit]

public class LongestCommonSubstring {
 
    public static void main(String[] args) {
        System.out.println(lcs("testing123testing", "thisisatest"));
        System.out.println(lcs("test", "thisisatest"));
        System.out.println(lcs("testing", "sting"));
        System.out.println(lcs("testing", "thisisasting"));
    }
 
    static String lcs(String a, String b) {
        if (a.length() > b.length())
            return lcs(b, a);
 
        String res = "";
        for (int ai = 0; ai < a.length(); ai++) {
            for (int len = a.length() - ai; len > 0; len--) {
 
                for (int bi = 0; bi <= b.length() - len; bi++) {
 
                    if (a.regionMatches(ai, b, bi, len) && len > res.length()) {
                        res = a.substring(ai, ai + len);
                    }
                }
            }
        }
        return res;
    }
}
test
test
sting
sting

JavaScript[edit]

Translation of: Haskell
(() => {
    'use strict';
 
    // longestCommon :: String -> String -> String
    const longestCommon = (s1, s2) => maximumBy(
        comparing(length),
        intersect(...apList(
            [s => map(
                concat,
                concatMap(tails, compose(tail, inits)(s))
            )],
            [s1, s2]
        ))
    );
 
    // main :: IO ()
    const main = () =>
        console.log(
            longestCommon(
                "testing123testing",
                "thisisatest"
            )
        );
 
    // GENERIC FUNCTIONS ----------------------------
 
    // Each member of a list of functions applied to each
    // of a list of arguments, deriving a list of new values.
 
    // apList (<*>) :: [(a -> b)] -> [a] -> [b]
    const apList = (fs, xs) => //
        fs.reduce((a, f) => a.concat(
            xs.reduce((a, x) => a.concat([f(x)]), [])
        ), []);
 
    // comparing :: (a -> b) -> (a -> a -> Ordering)
    const comparing = f =>
        (x, y) => {
            const
                a = f(x),
                b = f(y);
            return a < b ? -1 : (a > b ? 1 : 0);
        };
 
    // compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
    const compose = (f, g) => x => f(g(x));
 
    // concat :: [[a]] -> [a]
    // concat :: [String] -> String
    const concat = xs =>
        0 < xs.length ? (() => {
            const unit = 'string' !== typeof xs[0] ? (
                []
            ) : '';
            return unit.concat.apply(unit, xs);
        })() : [];
 
    // concatMap :: (a -> [b]) -> [a] -> [b]
    const concatMap = (f, xs) =>
        xs.reduce((a, x) => a.concat(f(x)), []);
 
    // inits([1, 2, 3]) -> [[], [1], [1, 2], [1, 2, 3]
    // inits('abc') -> ["", "a", "ab", "abc"]
 
    // inits :: [a] -> [[a]]
    // inits :: String -> [String]
    const inits = xs => [
            []
        ]
        .concat(('string' === typeof xs ? xs.split('') : xs)
            .map((_, i, lst) => lst.slice(0, i + 1)));
 
    // intersect :: (Eq a) => [a] -> [a] -> [a]
    const intersect = (xs, ys) =>
        xs.filter(x => -1 !== ys.indexOf(x));
 
    // Returns Infinity over objects without finite length.
    // This enables zip and zipWith to choose the shorter
    // argument when one is non-finite, like cycle, repeat etc
 
    // length :: [a] -> Int
    const length = xs =>
        (Array.isArray(xs) || 'string' === typeof xs) ? (
            xs.length
        ) : Infinity;
 
    // map :: (a -> b) -> [a] -> [b]
    const map = (f, xs) => xs.map(f);
 
    // maximumBy :: (a -> a -> Ordering) -> [a] -> a
    const maximumBy = (f, xs) =>
        0 < xs.length ? (
            xs.slice(1)
            .reduce((a, x) => 0 < f(x, a) ? x : a, xs[0])
        ) : undefined;
 
    // tail :: [a] -> [a]
    const tail = xs => 0 < xs.length ? xs.slice(1) : [];
 
    // tails :: [a] -> [[a]]
    const tails = xs => {
        const
            es = ('string' === typeof xs) ? (
                xs.split('')
            ) : xs;
        return es.map((_, i) => es.slice(i))
            .concat([
                []
            ]);
    };
 
    // MAIN ---
    return main();
})();
Output:
test

jq[edit]

Translation of: C#, Go, Ruby
Works with: jq version 1.4

Utility functions: 

# Create an m x n matrix
def matrix(m; n; init):
  if m == 0 then []
  elif m == 1 then [range(0;n) | init]
  elif m > 0 then
    matrix(1;n;init) as $row
    | [range(0;m) | $row ]
  else error("matrix\(m);_;_) invalid")
  end;
 
def set(i;j; value):
  setpath([i,j]; value);

Longest Common Substring: 

def lcs(a; b):
  matrix(a|length; b|length; 0) as $lengths
  # state: [ $lengths, greatestLength, answer ]
  | [$lengths, 0]
  | reduce range(0; a|length) as $i 
      (.;
       reduce range(0; b|length) as $j 
         (.;
           if a[$i:$i+1] == b[$j:$j+1] then
            (if $i == 0 or $j == 0 then 1
             else .[0][$i-1][$j-1] + 1
 	     end) as $x
            | .[0] |= set($i; $j; $x)
            | if $x > .[1] then
                .[1] = $x
                | .[2] = a[1+$i - $x : 1+$i] # output
              else .
              end
          else .
          end )) | .[2];

Example: 

lcs("thisisatest"; "testing123testing")
Output:
$ jq -n -f Longest_common_substring.jq
"test"

Julia[edit]

Works with: Julia version 1.5
function lcs(s1::AbstractString, s2::AbstractString)
    l, r = 1, 0
    sub_len = 0
    for i in 1:length(s1)
        for j in i:length(s1)
            if !contains(s2, SubString(s1, i, j)) break
            elseif sub_len < j - i
                l, r = i, j
                sub_len = j - i
            end
        end
    end 
    s1[l:r] 
end
 
@show lcs("thisisatest", "testing123testing")

Kotlin[edit]

Translation of: Java
// version 1.1.2
 
fun lcs(a: String, b: String): String {
    if (a.length > b.length) return lcs(b, a)
    var res = ""
    for (ai in 0 until a.length) {
        for (len in a.length - ai downTo 1) {
            for (bi in 0 until b.length - len) {
                if (a.regionMatches(ai, b, bi,len) && len > res.length) {
                    res = a.substring(ai, ai + len)
                }
            }
        }
    }
    return res
}
 
fun main(args: Array<String>) = println(lcs("testing123testing", "thisisatest"))
Output:
test

Lobster[edit]

Translation of: Go
import std
def lcs(a, b) -> string:
    var out = ""
    let lengths = map(a.length * b.length): 0
    var greatestLength = 0
    for(a) x, i:
        for(b) y, j:
            if x == y:
                if i == 0 or j == 0:
                    lengths[i * b.length + j] = 1
                else:
                    lengths[i * b.length + j] = lengths[(i-1) * b.length + j - 1] + 1
                if lengths[i * b.length + j] > greatestLength:
                    greatestLength = lengths[i * b.length + j]
                    out = a.substring(i - greatestLength + 1, greatestLength)
    return out
Translation of: C#
import std
def lcs2(a, b) -> string:
    var out = ""
    let lengths = map(b.length): map(a.length): 0
    var greatestLength = 0
    for(a) x, i:
        for(b) y, j:
            if x == y:
                if i == 0 or j == 0:
                    lengths[j][i] = 1
                else:
                    lengths[j][i] = lengths[j-1][i-1] + 1
                if lengths[j][i] > greatestLength:
                    greatestLength = lengths[j][i]
                    out = a.substring(i - greatestLength + 1, greatestLength)
    return out

Maple[edit]

StringTools:-LongestCommonSubString() returns the longest common substring of two strings. StringTools:-CommonSubSequence() returns the longest common subsequence() of two strings. 

StringTools:-LongestCommonSubString("thisisatest","testing123testing");

Mathematica/Wolfram Language[edit]

The function LongestCommonSubsequence returns the longest common substring, and LongestCommonSequence returns the longest common subsequence. 

Print[LongestCommonSubsequence["thisisatest", "testing123testing"]];
Output:
test

Modula-2[edit]

MODULE LCS;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,WriteLn,Write,ReadChar;
 
PROCEDURE WriteSubstring(s : ARRAY OF CHAR; b,e : CARDINAL);
VAR i : CARDINAL;
BEGIN
    IF b=e THEN RETURN END;
    IF e>HIGH(s) THEN e := HIGH(s) END;
 
    FOR i:=b TO e DO
        Write(s[i])
    END
END WriteSubstring;
 
TYPE
    Pair = RECORD
        a,b : CARDINAL;
    END;
 
PROCEDURE lcs(sa,sb : ARRAY OF CHAR) : Pair;
VAR
    output : Pair;
    a,b,len : CARDINAL;
BEGIN
    output := Pair{0,0};
 
    FOR a:=0 TO HIGH(sa) DO
        FOR b:=0 TO HIGH(sb) DO
            IF (sa[a]#0C) AND (sb[b]#0C) AND (sa[a]=sb[b]) THEN
                len := 1;
                WHILE (a+len<HIGH(sa)) AND (b+len<HIGH(sb)) DO
                    IF sa[a+len] = sb[b+len] THEN
                        INC(len)
                    ELSE
                        BREAK
                    END
                END;
                DEC(len);
 
                IF len>output.b-output.a THEN
                    output := Pair{a,a+len}
                END
            END
        END
    END;
 
    RETURN output
END lcs;
 
VAR res : Pair;
BEGIN
    res := lcs("testing123testing", "thisisatest");
    WriteSubstring("testing123testing", res.a, res.b);
    WriteLn;
 
    ReadChar
END LCS.

Nim[edit]

Translation of: Go
# Longest common substring.
 
import sequtils
 
func lcs(a, b: string): string =
  var lengths = newSeqWith(a.len, newSeq[int](b.len))
  var greatestLength = 0
  for i, x in a:
    for j, y in b:
      if x == y:
        lengths[i][j] = if i == 0 or j == 0: 1 else: lengths[i - 1][j - 1] + 1
        if lengths[i][j] > greatestLength:
          greatestLength = lengths[i][j]
          result = a[(i - greatestLength + 1)..i]
 
echo lcs("thisisatest", "testing123testing")
Output:
test

Pascal[edit]

Translation of: Java

using FreePascal[edit]

Works with: FreePascal version version 3.2.0
 
PROGRAM LongestCommonSubString.pas;
 
{$DEFINE DEBUGGING}
{$IFDEF FPC}
    {$mode objfpc}{$H+}{$J-}{R+}
{$ELSE}
    {$APPTYPE CONSOLE}
{$ENDIF}
 
(*)
 
        Free Pascal Compiler version 3.2.0 [2020/06/14] for x86_64
        The free and readable alternative at C/C++ speeds
        compiles natively to almost any platform, including raspberry PI *
        Can run independently from DELPHI / Lazarus
 
(*)
 
    FUNCTION lcss( S1, S2: string ) : string ;
 
    (*)
 
        LongestCommonSubString: plain version without extra libraries
 
        From https://en.wikibooks.org/wiki/Algorithm_implementation/Strings/Longest_common_substring
        translated from java
 
    (*)
 
        VAR
            i:      integer = 0;
            j:      integer = 0;
            x:      integer = 0;
            LenS1:  integer = 0;
            LenS2:  integer = 0;
            Start:  integer = 0;
            Max:    integer = 0;
 
        BEGIN
 
            LenS1 := length ( S1 ) - 1 ;
            LenS2 := length ( S2 ) - 1 ;
 
            FOR i := 0 TO  LenS1 DO
                BEGIN
 
                    FOR j := 0 TO LenS2 DO
                        BEGIN
 
                            x := 0;
 
                            WHILE ( S1 [ i + x ] = S2 [ j + x ] ) DO
 
                                BEGIN
                                    Inc ( x ) ;
                                    IF ( ( ( i + x ) > LenS1 ) or ( ( j + x ) > LenS2 ) ) THEN BREAK ;
                                END ;
 
                            IF ( x > Max ) THEN
 
                                BEGIN
                                    Max   := x ;
                                    Start := i ;
                                END ;
                        END ;
                END ;
 
            lcss :=  copy ( S1, Start, ( Start + Max ) ) ;
 
        END ;
 
VAR
    S1: string ;
    S2: string ;
 
BEGIN
 
    S1 :=   'thisisatest'       ;
    S2 :=   'testing123testing' ;
 
    WriteLn( Lcss ( S1, S2 ) )  ;
 
END.
 
 
JPD 2021/06/18

Output:

test

using FreePascal v.2[edit]

Works with: FreePascal version version 3.2.0
 
PROGRAM LongestCommonSubString.pas;
 
{$IFDEF FPC}
    {$mode objfpc}{$H+}{$J-}{R+}
{$ELSE}
    {$APPTYPE CONSOLE}
{$ENDIF}
 
(*)
 
        Free Pascal Compiler version 3.2.0 [2020/06/14] for x86_64
        The free and readable alternative at C/C++ speeds
        compiles natively to almost any platform, including raspberry PI *
        Can run independently from DELPHI / Lazarus
 
(*)
 
USES
 
    SysUtils;
 
    FUNCTION lcss( S1, S2: string ) : string ;
 
    (*)
 
        LongestCommonSubString: adaptation from Delphi
 
    (*)
 
    VAR
        j:          Integer =  0 ;
        k:          Integer =  0 ;
        lenS1:      integer =  0 ;
        S:          string  = '' ;
 
    BEGIN
 
        Result  := S ;
        lenS1   := S1.Length ;
 
        FOR j := 0 TO lenS1 - 1 DO
 
            BEGIN
                FOR k := lenS1 - j DOWNTO 0 DO
 
                    BEGIN
                        S := S1.Substring ( j, k ) ;
                        IF ( S2.IndexOf ( S ) > -1 ) AND ( S.Length > Result.Length ) THEN
                            Result := S;
                    END;
            END;
    END;
 
VAR
    S1: string ;
    S2: string ;
 
BEGIN
 
    S1 :=   'thisisatest'       ;
    S2 :=   'testing123testing' ;
 
  IF ParamCount = 2 THEN
    BEGIN
        IF NOT ParamStr(1).IsEmpty THEN
            S1 := ParamStr(1);
        IF NOT ParamStr(2).IsEmpty THEN
            S2 := ParamStr(2);
    END;
 
    Writeln ('string A = ', S1)  ;
    Writeln ('string B = ', S2)  ;
    WriteLn ( Lcss ( S1, S2 ) )  ;
 
END.    
 
JPD 2021/06/18

Output:

string A = thisisatest

string B = testing123testing

test

Perl[edit]

#!/usr/bin/perl
use strict ;
use warnings ;
 
sub longestCommonSubstr {
   my $first = shift ;
   my $second = shift ;
   my %firstsubs = findSubstrings ( $first );
   my %secondsubs = findSubstrings ( $second ) ;
   my @commonsubs ;
   foreach my $subst ( keys %firstsubs ) {
      if ( exists $secondsubs{ $subst } ) {
	 push ( @commonsubs , $subst ) ;
      }
   }
   my @sorted = sort { length $b <=> length $a } @commonsubs ;
   return $sorted[0] ;
}
 
sub findSubstrings {
   my $string = shift ;
   my %substrings ;
   my $l = length $string ;
   for ( my $start = 0 ; $start < $l ; $start++ ) {
      for ( my $howmany = 1 ; $howmany < $l - $start + 1 ; $howmany++) {
	 $substrings{substr( $string , $start , $howmany) } = 1 ;
      }
   }
   return %substrings ;
}
 
my $longest = longestCommonSubstr( "thisisatest" ,"testing123testing" ) ;
print "The longest common substring of <thisisatest> and <testing123testing> is $longest !\n" ; 
 
Output:
The longest common substring of <thisisatest> and <testing123testing> is test !

Alternate letting regex do the work[edit]

#!/usr/bin/perl
 
use strict; # https://rosettacode.org/wiki/Longest_Common_Substring
use warnings;
 
my $one = 'thisisatest';
my $two = 'testing123testing';
 
my @best;
"$one\n$two" =~ /(.+).*\n.*\1(?{ $best[length $1]{$1}++})(*FAIL)/;
print "$_\n" for sort keys %{ $best[-1] };
Output:

test

Phix[edit]

function lcs(string a, b)
integer longest = 0
string best = ""
    for i=1 to length(a) do
        integer ch = a[i]
        for j=1 to length(b) do
            if ch=b[j] then
                integer n=1
                while i+n<=length(a)
                  and j+n<=length(b)
                  and a[i+n]=b[j+n] do
                    n += 1
                end while
                if n>longest then
                    longest = n
                    best = a[i..i+n-1]
                end if
            end if
        end for
    end for
    return best
end function
?lcs("thisisatest", "testing123testing")
?lcs("testing123testing","thisisatest")
Output:
"test"
"test"

PicoLisp[edit]

(de longestCommonSubstring (Str1 Str2)
   (setq Str1 (chop Str1)  Str2 (chop Str2))
   (let Res NIL
      (map
         '((Lst1)
            (map
               '((Lst2)
                  (let Len 0
                     (find
                        '((A B) (nand (= A B) (inc 'Len)))
                        Lst1
                        Lst2 )
                     (when (> Len (length Res))
                        (setq Res (head Len Lst1)) ) ) )
               Str2 ) )
         Str1 )
      (pack Res) ) )

Test: 

: (longestCommonSubstring "thisisatest" "testing123testing")
-> "test"

PowerShell[edit]

function lcs([String]$a, [String]$b) 
{
    if ([String]::IsNullOrEmpty($a) -or [String]::IsNullOrEmpty($b)) 
    {
        return ""
    }
    $startIndex, $size = -1, -1
    for ($k = 0; $k -lt $a.Length; ++$k) 
    {
        for ($i, $j, $d = $k, 0, 0; ($i -lt $a.Length) -and ($j -lt $b.Length); ++$i, ++$j) 
        {
            if ($a.Chars($i) -eq $b.Chars($j))
            {
                $d += 1
                if ($size -lt $d)
                {
                    $startIndex = $i - $d + 1
                    $size = $d
                }
            } 
            else 
            {
                $d = 0
            }
        }
    }
    for ($k = 1; $k -lt $b.Length; ++$k) 
    {
        for ($i, $j, $d = 0, $k, 0; ($i -lt $a.Length) -and ($j -lt $b.Length); ++$i, ++$j) 
        {
            if ($a.Chars($i) -eq $b.Chars($j))
            {
                $d += 1
                if ($size -lt $d)
                {
                    $startIndex = $i - $d + 1
                    $size = $d
                }
            } 
            else 
            {
                $d = 0
            }
        }
    }
    if ($size -lt 0)
    {
        return ""
    }
    return $a.Substring($startIndex, $size)
}
 
function Print-Lcs([String]$a, [String]$b)  
{
    return "lcs $a $b = $(lcs $a $b)"
}
Print-Lcs 'thisisatest' 'testing123testing'
Print-Lcs 'testing' 'sting'
Print-Lcs 'thisisatest_stinger' 'testing123testingthing'
Print-Lcs 'thisisatest_stinger' 'thisis'
Print-Lcs 'testing123testingthing' 'thisis'
Print-Lcs 'thisisatest' 'thisisatest'
Output:
lcs thisisatest testing123testing = test
lcs testing sting = sting
lcs thisisatest_stinger testing123testingthing = sting
lcs thisisatest_stinger thisis = thisis
lcs testing123testingthing thisis = thi
lcs thisisatest thisisatest = thisisatest

Prolog[edit]

common_sublist(A, B, M) :-
	append(_, Ma, A), 
	append(M, _, Ma),
	append(_, Mb, B),
	append(M, _, Mb).	
 
longest_list([], L, _, L).
longest_list([L|Ls], LongestList, LongestLength, Result) :-	
	length(L, Len),
	Len >= LongestLength -> longest_list(Ls, L, Len, Result)
	; longest_list(Ls, LongestList, LongestLength, Result).
 
longest_substring(A, B, Result) :-
	string_chars(A, AChars),
	string_chars(B, BChars),
	findall(SubString, (
		dif(SubString, []), common_sublist(AChars, BChars, SubString)
	), AllSubstrings),
	longest_list(AllSubstrings, [], 0, LongestSubString),
	string_chars(Result, LongestSubString).
Output:
?- longest_substring("thisisatest", "testing123testing", Longest).
Longest = "test".

PureBasic[edit]

Procedure.s lcs(a$,b$)
  If Len(a$)>Len(b$) : Swap a$,b$ : EndIf  
  l=Len(a$)
  For c=1 To l
    For i=1 To 1+l-c      
      If FindString(b$,Mid(a$,i,c))
        res$=Mid(a$,i,c)
      EndIf
    Next
  Next
  ProcedureReturn res$
EndProcedure
 
t1$="testing123testing"
t2$="thisisatest"
Debug lcs(t1$,t2$)
Output:
test

Python[edit]

Python: Idiomatic[edit]

Python: Using Indexes[edit]

s1 = "thisisatest"
s2 = "testing123testing"
len1, len2 = len(s1), len(s2)
ir, jr = 0, -1
for i1 in range(len1):
    i2 = s2.find(s1[i1])
    while i2 >= 0:
        j1, j2 = i1, i2
        while j1 < len1 and j2 < len2 and s2[j2] == s1[j1]:
            if j1-i1 >= jr-ir:
                ir, jr = i1, j1
            j1 += 1; j2 += 1
        i2 = s2.find(s1[i1], i2+1)
print (s1[ir:jr+1])
Output:
"test"

Python: Set of substrings[edit]

From my explanatory blog post. 

def _set_of_substrings(s:str) -> set:
    "_set_of_substrings('ABBA') == {'A', 'AB', 'ABB', 'ABBA', 'B', 'BA', 'BB', 'BBA'}"
    len_s = len(s)
    return {s[m: n] for m in range(len_s) for n in range(m+1, len_s +1)}
 
def _set_of_common_substrings(s:str, common: set) -> set:
    "Substrings of s that are also in common"
    len_s = len(s)
    return {this for m in range(len_s) for n in range(m+1, len_s +1)
            if (this := s[m:n]) in common}
 
def lcs_ss(*strings):
    "longest of the common substrings of all substrings of each string"
    strings_iter  = (s for s in strings)
    common = _set_of_substrings(next(strings_iter)) # First string substrings
    for s in strings_iter:
        if not common:
            break
        common = _set_of_common_substrings(s, common) # Accumulate the common
 
    return max(common, key= lambda x: len(x)) if common else ''
 
 
s0 = "thisisatest_stinger"
s1 = "testing123testingthing"
s2 = "thisis"
 
ans = lcs_ss(s0, s1)
print(f"\n{repr(s0)}, {repr(s1)} ->> {repr(ans)}")
ans = lcs_ss(s0, s2)
print(f"\n{repr(s0)}, {repr(s2)} ->> {repr(ans)}")
ans = lcs_ss(s1, s2)
print(f"\n{repr(s1)}, {repr(s2)} ->> {repr(ans)}")
ans = lcs_ss(s0, s1, s2)
print(f"\n{repr(s0)}, {repr(s1)}, {repr(s2)} ->> {repr(ans)}")
 
Output:
'thisisatest_stinger', 'testing123testingthing' ->> 'sting'

'thisisatest_stinger', 'thisis' ->> 'thisis'

'testing123testingthing', 'thisis' ->> 'thi'

'thisisatest_stinger', 'testing123testingthing', 'thisis' ->> 'thi'


Functional[edit]

Translation of: Haskell
Translation of: JavaScript


Expressed as a composition of pure generic functions: 

'''Longest common substring'''
 
from itertools import accumulate, chain
from functools import reduce
 
 
# longestCommon :: String -> String -> String
def longestCommon(s1):
    '''The longest common substring of
       two given strings.
    '''
    def go(s2):
        return max(intersect(
            *map(lambda s: map(
                concat,
                concatMap(tails)(
                    compose(tail, list, inits)(s)
                )
            ), [s1, s2])
        ), key=len)
    return go
 
 
# ------------------------- TEST -------------------------
def main():
    '''Test'''
    print(
        longestCommon(
            "testing123testing"
        )(
            "thisisatest"
        )
    )
 
 
# ----------------------- GENERIC ------------------------
 
# compose :: ((a -> a), ...) -> (a -> a)
def compose(*fs):
    '''Composition, from right to left,
       of a series of functions.
    '''
    def go(f, g):
        def fg(x):
            return f(g(x))
        return fg
    return reduce(go, fs, lambda x: x)
 
 
# concat :: [String] -> String
def concat(xs):
    '''The concatenation of all the elements
       in a list or iterable.
    '''
    return ''.join(xs)
 
 
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
    '''A concatenated list over which a function has been
       mapped.
       The list monad can be derived by using a function f
       which wraps its output in a list, (using an empty
       list to represent computational failure).
    '''
    def go(xs):
        return chain.from_iterable(map(f, xs))
    return go
 
 
# inits :: [a] -> [[a]]
def inits(xs):
    '''all initial segments of xs, shortest first.'''
    return accumulate(chain([[]], xs), lambda a, x: a + [x])
 
 
# intersect :: [a] -> [a] -> [a]
def intersect(xs, ys):
    '''The ordered intersection of xs and ys.
       intersect([1,2,2,3,4])([6,4,4,2]) -> [2,2,4]
    '''
    s = set(ys)
    return (x for x in xs if x in s)
 
 
# scanl :: (b -> a -> b) -> b -> [a] -> [b]
def scanl(f):
    '''scanl is like reduce, but defines a succession of
       intermediate values, building from the left.
    '''
    def go(a):
        def g(xs):
            return accumulate(chain([a], xs), f)
        return g
    return go
 
 
# tail :: [a] -> [a]
# tail :: Gen [a] -> [a]
def tail(xs):
    '''The elements following the head of a
       (non-empty) list.
    '''
    return xs[1:]
 
 
# tails :: [a] -> [[a]]
def tails(xs):
    '''All final segments of xs,
       longest first.
    '''
    return [
        xs[i:] for i in
        range(0, 1 + len(xs))
    ]
 
 
# MAIN ---
main()
 
test

Racket[edit]

A chance to show off how to use HashTable types in typed/racket 

#lang typed/racket
(: lcs (String String -> String))
(define (lcs a b)
  (: all-substrings# (String -> (HashTable String Boolean)))
  (define (all-substrings# str)
    (define l (string-length str))
    (for*/hash : (HashTable String Boolean)
      ((s (in-range 0 l)) (e (in-range (add1 s) (add1 l))))
      (values (substring str s e) #t)))
 
  (define a# (all-substrings# a))
 
  (define b# (all-substrings# b))
 
  (define-values (s l)
    (for/fold : (Values String Nonnegative-Integer)
    ((s "") (l : Nonnegative-Integer 0))
    ((a_ (in-hash-keys a#))
     #:when (and (> (string-length a_) l) (hash-ref b# a_ #f)))
    (values a_ (string-length a_))))
 
  s)
 
(module+ test
  ("thisisatest" . lcs . "testing123testing"))
Output:
"test"

Raku[edit]

(formerly Perl 6) 

sub createSubstrings( Str $word --> Array ) {
   my $length = $word.chars ;
   my @substrings ;
   for (0..$length - 1) -> $start {
      for (1..$length - $start) -> $howmany {
	 @substrings.push( $word.substr( $start , $howmany ) ) ;
      }
   }
   return @substrings ;
}
 
sub findLongestCommon( Str $first , Str $second --> Str ) {
   my @substringsFirst  = createSubstrings( $first ) ;
   my @substringsSecond = createSubstrings( $second ) ;
   my $firstset  = set( @substringsFirst ) ;
   my $secondset = set( @substringsSecond ) ;
   my $common = $firstset (&) $secondset ;
   return $common.keys.sort({$^b.chars <=> $^a.chars})[0] ; # or: sort(-*.chars)[0]
}
 
sub MAIN( Str $first , Str $second ) {
   say "The longest common substring of $first and $second is " ~
   "{findLongestCommon( $first , $second ) } !" ;
}
Output:
The longest common substring of thisisatest and testing123testing is test !

Functional[edit]

sub substrings ($s)       { (flat (0..$_ X 1..$_).grep:{$_ ≥ [+] @_}).map: { $s.substr($^a, $^b) } given $s.chars }
sub infix:<LCS>($s1, $s2) { ([∩] ($s1, $s2)».&substrings).keys.sort(*.chars).tail }
 
my $first  = 'thisisatest';
my $second = 'testing123testing';
say "The longest common substring between '$first' and '$second' is '{$first LCS $second}'.";
Output:
The longest common substring between 'thisisatest' and 'testing123testing' is 'test'.

REXX[edit]

/*REXX program determines the   LCSUBSTR   (Longest Common Substring)  via a function.  */
parse arg a b .                                  /*obtain optional arguments from the CL*/
if a==''  then a= "thisisatest"                  /*Not specified?  Then use the default.*/
if b==''  then b= "testing123testing"            /* "      "         "   "     "    "   */
say '   string A ='            a                 /*echo string A to the terminal screen.*/
say '   string B ='               b              /*  "     "   B  "  "      "       "   */
say '   LCsubstr ='   LCsubstr(a, b)             /*display the Longest Common Substring.*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LCsubstr: procedure;  parse arg x,y,,$;     #= 0 /*LCsubstr:  Longest Common Substring. */
          L= length(x);     w= length(y)         /*placeholders for string length of X,Y*/
          if w<L  then do;  parse arg y,x;  L= w /*switch X & Y   if Y is shorter than X*/
                       end
             do j=1  for L  while j<=L-#         /*step through start points in string X*/
                do k=L-j+1   to #   by -1        /*step through string lengths.         */
                _= substr(x, j, k)               /*extract a possible common substring. */
                if pos(_, y)\==0  then  if k>#  then do;     $= _;     #= k;      end
                end   /*k*/                      /* [↑]  determine if string _ is longer*/
             end      /*j*/                      /*#:  the current length of  $  string.*/
          return $                               /*$:  (null if there isn't common str.)*/
output   when using the default inputs:
   string A = thisisatest
   string B = testing123testing
   LCsubstr = test

Ring[edit]

 
# Project : Longest Common Substring
 
str1 = "testing123testing" 
str2 = "tsitest"
see longest(str1, str2)
 
func  longest(str1, str2)
subarr = []
for n=1 to len(str1)
    for m=1 to len(str1)
        sub = substr(str1, n, m)
        if substr(str2, sub) > 0
           add(subarr, sub)
        ok
    next
next
 
temp = 0
for n=1 to len(subarr) 
    if len(subarr[n]) > temp
       temp = len(subarr[n])
       subend = subarr[n]
    ok
next
see subend + nl
 

Output: 

test

Ruby[edit]

Translation of: C#
def longest_common_substring(a,b)
  lengths = Array.new(a.length){Array.new(b.length, 0)}
  greatestLength = 0
  output = ""
  a.each_char.with_index do |x,i|
    b.each_char.with_index do |y,j|
      next if x != y
      lengths[i][j] = (i.zero? || j.zero?) ? 1 : lengths[i-1][j-1] + 1
      if lengths[i][j] > greatestLength
        greatestLength = lengths[i][j]
        output = a[i - greatestLength + 1, greatestLength]
      end
    end
  end
  output
end
 
p longest_common_substring("thisisatest", "testing123testing")
Output:
"test"

Rust[edit]

fn longest_common_substring(s1: &str, s2: &str) -> String {
    let s1_chars: Vec<char> = s1.chars().collect();
    let s2_chars: Vec<char> = s2.chars().collect();
    let mut lcs = "".to_string();
 
    for i in 0..s1_chars.len() {
        for j in 0..s2_chars.len() {
            if s1_chars[i] == s2_chars[j] {
                let mut tmp_lcs = s2_chars[j].to_string();
                let mut tmp_i = i + 1;
                let mut tmp_j = j + 1;
 
                while tmp_i < s1_chars.len() && tmp_j < s2_chars.len() && s1_chars[tmp_i] == s2_chars[tmp_j] {
                    tmp_lcs = format!("{}{}", tmp_lcs, s1_chars[tmp_i]);
                    tmp_i += 1;
                    tmp_j += 1;
                }
 
                if tmp_lcs.len() > lcs.len() {
                    lcs = tmp_lcs;
                }
            }
        }
    }
 
    lcs
}
 
fn main() {
    let s1 = "thisisatest";
    let s2 = "testing123testing";
    let lcs = longest_common_substring(s1, s2);
    println!("{}", lcs);
}
Output:
"test"

Scala[edit]

The two algorithms below are Scala optimized versions of the Dynamic Programming algorithm pseudocode solution found on the "Longest Common Substring" Wikipedia page.

For a more in-depth look at the Scala solution space for this problem, please see this StackOverflow answer.

longestCommonSubstringsOptimizedPureFP

This algorithm sticks to "Pure" FP (Functional Programming) in that it uses tail recursion while avoiding any use of mutable collections or vars within the function's implementation.

Explore and experiment withing the online Scala playgrounds that run in your favorite browser: ScalaFiddle (ES a.k.a. JavaScript, non JVM) or Scastie (remote JVM). 

def longestCommonSubstringsOptimizedPureFP(left: String, right: String): Option[Set[String]] =
  if (left.nonEmpty && right.nonEmpty) {
    val (shorter, longer) =
      if (left.length < right.length) (left, right)
      else (right, left)
 
    @scala.annotation.tailrec
    def recursive(
      indexLonger: Int = 0,
      indexShorter: Int = 0,
      currentLongestLength: Int = 0,
      lengthsPrior: List[Int] = List.fill(shorter.length)(0),
      lengths: List[Int] = Nil,
      accumulator: List[Int] = Nil
    ): (Int, List[Int]) =
      if (indexLonger < longer.length) {
        val length =
          if (longer(indexLonger) != shorter(indexShorter)) 0
          else lengthsPrior.head + 1
        val newCurrentLongestLength =
          if (length > currentLongestLength) length
          else currentLongestLength
        val newAccumulator =
          if ((length < currentLongestLength) || (length == 0)) accumulator
          else {
            val entry = indexShorter - length + 1
            if (length > currentLongestLength) List(entry)
            else entry :: accumulator
          }
        if (indexShorter < shorter.length - 1)
          recursive(
            indexLonger,
            indexShorter + 1,
            newCurrentLongestLength,
            lengthsPrior.tail,
            length :: lengths,
            newAccumulator
          )
        else
          recursive(
            indexLonger + 1,
            0,
            newCurrentLongestLength,
            0 :: lengths.reverse,
            Nil,
            newAccumulator
          )
      }
      else (currentLongestLength, accumulator)
 
    val (length, indexShorters) = recursive()
    if (indexShorters.nonEmpty)
      Some(
        indexShorters
          .map {
            indexShorter =>
              shorter.substring(indexShorter, indexShorter + length)
          }
          .toSet
      )
    else None
  }
  else None
 
println(longestCommonSubstringsOptimizedPureFP("thisisatest", "testing123testing"))
Output:
"Some(Set(test))"

longestCommonSubstringsOptimizedReferentiallyTransparentFP

While this algorithm remains consistent with the FP concept of referential transparency, it does use both a mutable collection and a var within the function's implementation which provide an almost three times performance improvement over the above longestCommonSubstringsOptimizedPureFP implementation.

Explore this visual diff to see the changes between the longestCommonSubstringsOptimizedPureFP (above) and longestCommonSubstringsOptimizedReferentiallyTransparentFP (below) implementations

Explore and experiment withing the online Scala playgrounds that run in your favorite browser: ScalaFiddle (ES a.k.a. JavaScript, non JVM) or Scastie (remote JVM). 

def longestCommonSubstringsOptimizedReferentiallyTransparentFP(left: String, right: String): Option[Set[String]] =
  if (left.nonEmpty && right.nonEmpty) {
    val (shorter, longer) =
      if (left.length < right.length) (left, right)
      else (right, left)
    val lengths: Array[Int] = new Array(shorter.length) //mutable
 
    @scala.annotation.tailrec
    def recursive(
      indexLonger: Int = 0,
      indexShorter: Int = 0,
      currentLongestLength: Int = 0,
      lastIterationLength: Int = 0,
      accumulator: List[Int] = Nil
    ): (Int, List[Int]) =
      if (indexLonger < longer.length) {
        val length =
          if (longer(indexLonger) != shorter(indexShorter)) 0
          else
            if (indexShorter == 0) 1
            else lastIterationLength + 1
        val newLastIterationLength = lengths(indexShorter)
        lengths(indexShorter) = length //mutation
        val newCurrentLongestLength =
          if (length > currentLongestLength) length
          else currentLongestLength
        val newAccumulator =
          if ((length < currentLongestLength) || (length == 0)) accumulator
          else {
            val entry = indexShorter - length + 1
            if (length > currentLongestLength) List(entry)
            else entry :: accumulator
          }
        if (indexShorter < shorter.length - 1)
          recursive(
            indexLonger,
            indexShorter + 1,
            newCurrentLongestLength,
            newLastIterationLength,
            newAccumulator
          )
        else
          recursive(
            indexLonger + 1,
            0,
            newCurrentLongestLength,
            newLastIterationLength,
            newAccumulator
          )
      }
      else (currentLongestLength, accumulator)
 
    val (length, indexShorters) = recursive()
    if (indexShorters.nonEmpty)
      Some(
        indexShorters
          .map {
            indexShorter =>
              shorter.substring(indexShorter, indexShorter + length)
          }
          .toSet
      )
    else None
  }
  else None
 
println(longestCommonSubstringsOptimizedReferentiallyTransparentFP("thisisatest", "testing123testing"))
Output:
"Some(Set(test))"

Sidef[edit]

Translation of: Raku
func createSubstrings(String word) -> Array {
  gather {
    combinations(word.len+1, 2, {|i,j|
        take(word.substr(i, j-i))
    })
  }
}
 
func findLongestCommon(String first, String second) -> String {
    createSubstrings(first) & createSubstrings(second) -> max_by { .len }
}
 
say findLongestCommon("thisisatest", "testing123testing")
Output:
test

Swift[edit]

func lComSubStr<
  S0: Sliceable, S1: Sliceable, T: Equatable where
  S0.Generator.Element == T, S1.Generator.Element == T,
  S0.Index.Distance == Int, S1.Index.Distance == Int
  >(w1: S0, _ w2: S1) -> S0.SubSlice {
 
    var (len, end) = (0, 0)
 
    let empty = Array(Repeat(count: w2.count + 1, repeatedValue: 0))
    var mat: [[Int]] = Array(Repeat(count: w1.count + 1, repeatedValue: empty))
 
    for (i, sLett) in w1.enumerate() {
      for (j, tLett) in w2.enumerate() where tLett == sLett {
        let curLen = mat[i][j] + 1
        mat[i + 1][j + 1] = curLen
        if curLen > len {
          len = curLen
          end = i
        }
      }
    }
    return w1[advance(w1.startIndex, (end + 1) - len)...advance(w1.startIndex, end)]
}
 
func lComSubStr(w1: String, _ w2: String) -> String {
  return String(lComSubStr(w1.characters, w2.characters))
}

Output: 

lComSubStr("thisisatest", "testing123testing") // "test"

VBA[edit]

 
Function Longest_common_substring(string1 As String, string2 As String) As String
Dim i As Integer, j As Integer, temp As String, result As String
    For i = 1 To Len(string1)
        For j = 1 To Len(string1)
            temp = Mid(string1, i, j)
            If InStr(string2, temp) Then
                If Len(temp) > Len(result) Then result = temp
            End If
        Next
    Next
    Longest_common_substring = result
End Function
 
Sub MainLCS()
    Debug.Print Longest_common_substring("thisisatest", "testing123testing")
End Sub
 
Output:

Invoke the script calling "MainLCS". 

test

VBScript[edit]

 
Function lcs(string1,string2)
	For i = 1 To Len(string1)
		tlcs = tlcs & Mid(string1,i,1)
		If InStr(string2,tlcs) Then
			If Len(tlcs) > Len(lcs) Then
				lcs = tlcs
			End If
		Else
			tlcs = ""
		End If
	Next
End Function
 
WScript.Echo lcs(WScript.Arguments(0),WScript.Arguments(1))
 
Output:

Invoke the script from a command prompt. 

C:\>cscript.exe /nologo lcs.vbs "thisisatest" "testing123testing"
test

Wren[edit]

Translation of: Go
var lcs = Fn.new { |a, b|
    var la = a.count
    var lb = b.count
    var lengths = List.filled(la * lb, 0)
    var greatestLength = 0
    var output = ""
    var i = 0
    for (x in a) {
        var j = 0
        for (y in b) {
            if (x == y) {
                lengths[i*lb + j] = (i == 0 || j == 0) ? 1 : lengths[(i-1)*lb+j-1] + 1
            }
            if (lengths[i*lb+j] > greatestLength) {
                greatestLength = lengths[i*lb+j]
                output = a[i-greatestLength+1..i]
            }
            j = j + 1
        }
        i = i + 1
    }
    return output
}
 
System.print(lcs.call("thisisatest", "testing123testing"))
Output:
test


Yabasic[edit]

Translation of: FreeBASIC
sub LCS$(a$, b$)
    if len(a$) = 0 or len(b$) = 0 then return "" : endif
	while len(b$)
		for j = len(b$) to 1 step -1
			if instr(a$, left$(b$, j)) then return left$(b$, j) : endif
		next j
		b$ = mid$(b$, 2)
	wend
end sub
 
print LCS$("thisisatest", "testing123testing")
end


zkl[edit]

fcn lcd(a,b){
   if(b.len()<a.len()){ t:=a; a=b; b=t; }
   foreach n,m in ([a.len()..1,-1],a.len()-n+1){
      s:=a[m,n];
      if(b.holds(s)) return(s);
   }
   ""
}
lcd("testing123testing","thisisatest").println();
Output:
test
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