Longest Common Substring

From Rosetta Code
Longest Common Substring is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Write a function 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".

References:

Metrics: length

Sub-string search: Count occurrences of a substring

Multi-string operations: LCP, LCS, concatenation

Manipulation: reverse, lower- and uppercase

Aime[edit]

void
test_string(text &g, text v, text l)
{
integer n;
 
n = 0;
while (l[n] && v[n] == l[n]) {
n += 1;
}
if (length(g) < n) {
g = cut(l, 0, n);
}
}
 
text
longest(text u, text v)
{
record r;
text g, l, s;
 
while (length(u)) {
r[u] = 0;
u = delete(u, 0);
}
while (length(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);
}
 
return g;
}
o_(longest("thisisatest", "testing123testing"), "\n");

AppleScript[edit]

on LCS(a as text, b as text)
local t1, t2
 
script
property s : substrings(a)
property t : substrings(b)
property list : missing value
 
on longest from L at w : ""
if length of L = 1 then return w
 
tell item 1 of L to if ¬
w's length < its length ¬
then set w to it
 
longest from the rest of L at w
end longest
end script
 
tell the result
repeat with x in (a reference to its s)
if x is not in its t then set the contents of x to null
end repeat
 
set its list to every string in its s
 
longest from its list
end tell
end LCS
 
 
on substrings(t)
local t
 
script
property result : {}
 
to iterate thru s at n : 1
local s, n
 
if the length of s < n then return {}
set my result to my result & (text 1 thru n of s)
iterate thru s at n + 1
end iterate
 
to recurse thru s
local s
 
if length of s = 1 then return s
iterate thru s
my result & substrings(text 2 thru -1 of s)
end recurse
end script
 
tell the result to recurse thru t
end substrings
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


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]

#include <string>
#include <algorithm>
#include <iostream>
#include <set>
#include <vector>
 
void findSubstrings ( const std::string & word , std::set<std::string> & substrings ) {
int l = word.length( ) ;
for ( int start = 0 ; start < l ; start++ ) {
for ( int length = 1 ; length < l - start + 1 ; length++ ) {
substrings.insert ( word.substr( start , length ) ) ;
}
}
}
 
std::string lcs ( const std::string & first , const std::string & second ) {
std::set<std::string> firstSubstrings , secondSubstrings ;
findSubstrings ( first , firstSubstrings ) ;
findSubstrings ( second , secondSubstrings ) ;
std::set<std::string> common ;
std::set_intersection ( firstSubstrings.begin( ) , firstSubstrings.end( ) ,
secondSubstrings.begin( ) , secondSubstrings.end( ) ,
std::inserter ( common , common.begin( ) ) ) ;
std::vector<std::string> commonSubs ( common.begin( ) , common.end( ) ) ;
std::sort ( commonSubs.begin( ) , commonSubs.end( ) , []( const std::string &s1 ,
const std::string &s2 ) { return s1.length( ) > s2.length( ) ; } ) ;
return *(commonSubs.begin( ) ) ;
}
 
int main( ) {
std::string s1 ("thisisatest" ) ;
std::string s2 ( "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 !

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

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

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

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 :: String -> [String]
subStrings s =
let intChars = length s
in [ take n $ drop i s
| i <- [0 .. intChars - 1]
, n <- [1 .. intChars - i] ]
 
longestCommon :: String -> String -> String
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 Data.Ord (comparing)
import Control.Monad ((<=<))
import Data.List (inits, intersect, maximumBy, tails)
 
longestCommon :: String -> String -> String
longestCommon a b =
maximumBy (comparing length) $
(uncurry intersect . pair) $ [tail . inits <=< tails] <*> [a, b]
 
pair :: [a] -> (a, a)
pair [x, y] = (x, y)
 
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

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

Java[edit]

public class LongestCommonSubstring {
 
public static void main(String[] args) {
System.out.println(lcs("testing123testing", "thisisatest"));
}
 
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

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 0.6
function lcs(s1::AbstractString, s2::AbstractString)
r = ""
i = 1
for i in 1:length(s1)
j = i
while j ≤ length(s1) && contains(s2, s1[i:j])
if length(r) < j - i + 1 r = s1[i:j] end
j += 1
end
end
return 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

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[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.

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 !

Perl 6[edit]

 
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] ;
}
 
sub MAIN( Str $first , Str $second ) {
my $phrase = "The longest common substring of $first and $second is " ~
"{findLongestCommon( $first , $second ) } !" ;
$phrase.say ;
}
Output:
The longest common substring of thisisatest and testing123testing is 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"

Python[edit]

Using Indexes[edit]

s1 = "thisisatest"
s2 = "testing123testing"
len1, len2 = len(s1), len(s2)
ir, jr = 0, 0
for i1 in range(len1):
i2 = s2.find(s1[i1])
while i2 >= 0:
j1, j2 = i1+1, i2+1
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"

Functional[edit]

Translation of: Haskell
Translation of: JavaScript


Expressed as a composition of generic functions:

from itertools import (accumulate, chain)
from functools import (reduce)
 
 
# longestCommon :: String -> String -> String
def longestCommon(s1):
return lambda s2: max(intersect(
*map(lambda s: map(
concat,
concatMap(tails)(
compose(tail)(inits)(s)
)
), [s1, s2])
), key=len)
 
 
# TEST ----------------------------------------------------
def main():
print(
longestCommon("testing123testing")(
"thisisatest"
)
)
 
 
# GENERIC -------------------------------------------------
 
 
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
return lambda f: lambda x: g(f(x))
 
 
# concat :: [String] -> String
def concat(xs):
return ''.join(chain.from_iterable(xs))
 
 
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
return lambda xs: list(
chain.from_iterable(
map(f, xs)
)
)
 
 
# inits :: [a] -> [[a]]
def inits(xs):
return scanl(lambda a, x: a + [x])(
[]
)(list(xs))
 
 
# intersect :: [a] -> [a] -> [a]
def intersect(xs, ys):
s = set(ys)
return [x for x in xs if x in s]
 
 
# map :: (a -> b) -> [a] -> [b]
def map_(f):
return lambda xs: list(map(f, xs))
 
 
# scanl is like reduce, but returns a succession of
# intermediate values, building from the left.
# scanl :: (b -> a -> b) -> b -> [a] -> [b]
 
 
def scanl(f):
return lambda a: lambda xs: (
list(accumulate([a] + list(xs), f))
)
 
 
# tail :: [a] -> [a]
def tail(xs):
return xs[1:]
 
 
# tails :: [a] -> [[a]]
def tails(xs):
return list(map(
lambda i: xs[i:],
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"

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 /*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"

Scala[edit]

Dynamic Programming[edit]

Functional Prog, (tail) recursive[edit]

Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
import scala.annotation.tailrec
 
object LongestCommonSubstring extends App {
 
def longestCommonSubstring(s: String, t: String): Seq[String] = {
val p = (s.length, t.length)
val nonEmpty = s.nonEmpty && t.nonEmpty
 
@tailrec
def iter(lcSufx: Map[(Int, Int), Int], indexes: (Int, Int), z: Int): Map[(Int, Int), Int] = {
val (i, j) = indexes
 
def newIndexes: (Int, Int) = if (j == p._2) (i + 1, 1) else (i, j + 1)
 
if (indexes != p && nonEmpty)
if (s(i - 1) == t(j - 1)) {
val count = lcSufx.withDefaultValue(0)((i - 1, j - 1)) + 1
 
@inline
def newLcSufx = lcSufx.filter(_._2 >= z).updated(indexes, count)
 
iter(newLcSufx, newIndexes, if (count >= z) count else z)
} else iter(lcSufx, newIndexes, z)
else lcSufx.filter(_._2 > 1)
}
 
iter(Map.empty[(Int, Int), Int], (1, 1), 0).map {
case ((i, _), z) => s.substring(i - z, i)
}.toSeq
}
 
println(longestCommonSubstring("testing123testing", "123thisisatest"))
 
}

Sidef[edit]

Translation of: Perl 6
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"

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

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