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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

{{task}}[[Category:Recursion]][[Category:Memoization]] The '''longest common subsequence''' (or [http://en.wikipedia.org/wiki/Longest_common_subsequence_problem '''LCS''']) of groups A and B is the longest group of elements from A and B that are common between the two groups and in the same order in each group. For example, the sequences "1234" and "1224533324" have an LCS of "1234": '''1234''' '''12'''245'''3'''332'''4''' For a string example, consider the sequences "thisisatest" and "testing123testing". An LCS would be "tsitest": '''t'''hi'''si'''sa'''test''' '''t'''e'''s'''t'''i'''ng123'''test'''ing

In this puzzle, your code only needs to deal with strings. Write a function which returns an LCS of two strings (case-sensitive). You don't need to show multiple LCS's.

Using recursion:

```with Ada.Text_IO;  use Ada.Text_IO;

procedure Test_LCS is
function LCS (A, B : String) return String is
begin
if A'Length = 0 or else B'Length = 0 then
return "";
elsif A (A'Last) = B (B'Last) then
return LCS (A (A'First..A'Last - 1), B (B'First..B'Last - 1)) & A (A'Last);
else
declare
X : String renames LCS (A, B (B'First..B'Last - 1));
Y : String renames LCS (A (A'First..A'Last - 1), B);
begin
if X'Length > Y'Length then
return X;
else
return Y;
end if;
end;
end if;
end LCS;
begin
Put_Line (LCS ("thisisatest", "testing123testing"));
end Test_LCS;
```

{{out}}

```
tsitest

```

Non-recursive solution:

```with Ada.Text_IO;  use Ada.Text_IO;

procedure Test_LCS is
function LCS (A, B : String) return String is
L : array (A'First..A'Last + 1, B'First..B'Last + 1) of Natural;
begin
for I in L'Range (1) loop
L (I, B'First) := 0;
end loop;
for J in L'Range (2) loop
L (A'First, J) := 0;
end loop;
for I in A'Range loop
for J in B'Range loop
if A (I) = B (J) then
L (I + 1, J + 1) := L (I, J) + 1;
else
L (I + 1, J + 1) := Natural'Max (L (I + 1, J), L (I, J + 1));
end if;
end loop;
end loop;
declare
I : Integer := L'Last (1);
J : Integer := L'Last (2);
R : String (1..Integer'Max (A'Length, B'Length));
K : Integer := R'Last;
begin
while I > L'First (1) and then J > L'First (2) loop
if L (I, J) = L (I - 1, J) then
I := I - 1;
elsif L (I, J) = L (I, J - 1) then
J := J - 1;
else
I := I - 1;
J := J - 1;
R (K) := A (I);
K := K - 1;
end if;
end loop;
return R (K + 1..R'Last);
end;
end LCS;
begin
Put_Line (LCS ("thisisatest", "testing123testing"));
end Test_LCS;
```

{{out}}

```
tsitest

```

## ALGOL 68

{{trans|Ada}} {{works with|ALGOL 68|Standard - no extensions to language used}} {{works with|ALGOL 68G|Any - tested with release mk15-0.8b.fc9.i386}} {{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386}}

```main:(
PROC lcs = (STRING a, b)STRING:
BEGIN
IF UPB a = 0 OR UPB b = 0 THEN
""
ELIF a [UPB a] = b [UPB b] THEN
lcs (a [:UPB a - 1], b [:UPB b - 1]) + a [UPB a]
ELSE
STRING x = lcs (a, b [:UPB b - 1]);
STRING y = lcs (a [:UPB a - 1], b);
IF UPB x > UPB y THEN x ELSE y FI
FI
END # lcs #;
print((lcs ("thisisatest", "testing123testing"), new line))
)
```

{{out}}

```
tsitest

```

## APL

{{works with|Dyalog APL}}

```lcs←{
⎕IO←0
betterof←{⊃(</+/¨⍺ ⍵)⌽⍺ ⍵}                     ⍝ better of 2 selections
cmbn←{↑,⊃∘.,/(⊂⊂⍬),⍵}                          ⍝ combine lists
rr←{∧/↑>/1 ¯1↓[1]¨⊂⍵}                          ⍝ rising rows
hmrr←{∨/(rr ⍵)∧∧/⍵=⌈\⍵}                        ⍝ has monotonically rising rows
rnbc←{{⍵/⍳⍴⍵}¨↓[0]×⍵}                          ⍝ row numbers by column
valid←hmrr∘cmbn∘rnbc                           ⍝ any valid solutions?
a w←(</⊃∘⍴¨⍺ ⍵)⌽⍺ ⍵                            ⍝ longest first
matches←a∘.=w
aps←{⍵[;⍒+⌿⍵]}∘{(⍵/2)⊤⍳2*⍵}                    ⍝ all possible subsequences
swps←{⍵/⍨∧⌿~(~∨⌿⍺)⌿⍵}                          ⍝ subsequences with possible solns
sstt←matches swps aps⊃⍴w                       ⍝ subsequences to try
w/⍨{
⍺←0⍴⍨⊃⍴⍵                                   ⍝ initial selection
(+/⍺)≥+/⍵[;0]:⍺                            ⍝ no scope to improve
this←⍺ betterof{⍵×valid ⍵/matches}⍵[;0]    ⍝ try to improve
1=1⊃⍴⍵:this                                ⍝ nothing left to try
this ∇ 1↓[1]⍵                              ⍝ keep looking
}sstt
}
```

## AutoHotkey

{{trans|Java}} using dynamic programming

ahk forum: [http://www.autohotkey.com/forum/viewtopic.php?t=44657&start=135 discussion]

```lcs(a,b) { ; Longest Common Subsequence of strings, using Dynamic Programming
Loop % StrLen(a)+2 {                          ; Initialize
i := A_Index-1
Loop % StrLen(b)+2
j := A_Index-1, len%i%_%j% := 0
}
Loop Parse, a                                 ; scan a
{
i := A_Index, i1 := i+1, x := A_LoopField
Loop Parse, b                              ; scan b
{
j := A_Index, j1 := j+1, y := A_LoopField
len%i1%_%j1% := x=y ? len%i%_%j% + 1
: (u:=len%i1%_%j%) > (v:=len%i%_%j1%) ? u : v
}
}
x := StrLen(a)+1, y := StrLen(b)+1
While x*y {                                   ; construct solution from lengths
x1 := x-1, y1 := y-1
If (len%x%_%y% = len%x1%_%y%)
x := x1
Else If  (len%x%_%y% = len%x%_%y1%)
y := y1
Else
x := x1, y := y1, t := SubStr(a,x,1) t
}
Return t
}
```

## BASIC

{{works with|QuickBasic|4.5}} {{trans|Java}}

```FUNCTION lcs\$ (a\$, b\$)
IF LEN(a\$) = 0 OR LEN(b\$) = 0 THEN
lcs\$ = ""
ELSEIF RIGHT\$(a\$, 1) = RIGHT\$(b\$, 1) THEN
lcs\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), LEFT\$(b\$, LEN(b\$) - 1)) + RIGHT\$(a\$, 1)
ELSE
x\$ = lcs\$(a\$, LEFT\$(b\$, LEN(b\$) - 1))
y\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), b\$)
IF LEN(x\$) > LEN(y\$) THEN
lcs\$ = x\$
ELSE
lcs\$ = y\$
END IF
END IF
END FUNCTION
```

## BBC BASIC

This makes heavy use of BBC BASIC's shortcut '''LEFT\$(a\$)''' and '''RIGHT\$(a\$)''' functions.

```      PRINT FNlcs("1234", "1224533324")
PRINT FNlcs("thisisatest", "testing123testing")
END

DEF FNlcs(a\$, b\$)
IF a\$="" OR b\$="" THEN = ""
IF RIGHT\$(a\$) = RIGHT\$(b\$) THEN = FNlcs(LEFT\$(a\$), LEFT\$(b\$)) + RIGHT\$(a\$)
LOCAL x\$, y\$
x\$ = FNlcs(a\$, LEFT\$(b\$))
y\$ = FNlcs(LEFT\$(a\$), b\$)
IF LEN(y\$) > LEN(x\$) SWAP x\$,y\$
= x\$
```

'''Output:'''

```
1234
tsitest

```

## Bracmat

```  ( LCS
=   A a ta B b tb prefix
.     !arg:(?prefix.@(?A:%?a ?ta).@(?B:%?b ?tb))
& ( !a:!b&LCS\$(!prefix !a.!ta.!tb)
| LCS\$(!prefix.!A.!tb)&LCS\$(!prefix.!ta.!B)
)
| !prefix:? ([>!max:[?max):?lcs
|
)
& 0:?max
& :?lcs
& LCS\$(.thisisatest.testing123testing)
& out\$(max !max lcs !lcs);
```

{{out}}

```max 7 lcs t s i t e s t
```

## C

```#include <stdio.h>
#include <stdlib.h>

#define MAX(a, b) (a > b ? a : b)

int lcs (char *a, int n, char *b, int m, char **s) {
int i, j, k, t;
int *z = calloc((n + 1) * (m + 1), sizeof (int));
int **c = calloc((n + 1), sizeof (int *));
for (i = 0; i <= n; i++) {
c[i] = &z[i * (m + 1)];
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= m; j++) {
if (a[i - 1] == b[j - 1]) {
c[i][j] = c[i - 1][j - 1] + 1;
}
else {
c[i][j] = MAX(c[i - 1][j], c[i][j - 1]);
}
}
}
t = c[n][m];
*s = malloc(t);
for (i = n, j = m, k = t - 1; k >= 0;) {
if (a[i - 1] == b[j - 1])
(*s)[k] = a[i - 1], i--, j--, k--;
else if (c[i][j - 1] > c[i - 1][j])
j--;
else
i--;
}
free(c);
free(z);
return t;
}

```

Testing

```int main () {
char a[] = "thisisatest";
char b[] = "testing123testing";
int n = sizeof a - 1;
int m = sizeof b - 1;
char *s = NULL;
int t = lcs(a, n, b, m, &s);
printf("%.*s\n", t, s); // tsitest
return 0;
}
```

## C++

'''The Longest Common Subsequence (LCS) Problem'''

Defining a subsequence to be a string obtained by deleting zero or more symbols from an input string, the LCS Problem is to find a subsequence of maximum length that is common to two input strings.

'''Background'''

Where the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols necessarily increases; and the number of matches will tend towards quadratic, O(m*n) growth.

This occurs, for example, in Bioinformatics applications of nucleotide and protein sequencing.

Here the "divide and conquer" approach of Hirschberg limits the space required to O(m+n). However, this approach requires O(m*n) time even in the best case.

This quadratic time dependency may become prohibitive, given very long input strings. Thus, heuristics are often favored over optimal Dynamic Programming solutions.

In the application of comparing file revisions, records from the input files form a large symbol space; and the number of symbols used for matches may approach the length of the LCS.

Assuming a uniform distribution of symbols, the number of matches may tend only towards linear, O(m+n) growth.

A binary search optimization due to Hunt and Szymanski can be applied in this case, which results in expected performance of O(n log m), given m <= n. In the worst case, performance degrades to O(mn log m) time if the number of matches, and the space required to represent them, should grow to O(mn).

More recent improvements by Rick and by Goeman and Clausen reduce the time bound to O(ns + min(pm, p*(n-p))) where the alphabet is of size s and the LCS is of length p.

'''References'''

"A linear space algorithm for computing maximal common subsequences"
by Daniel S. Hirschberg, published June 1975
Communications of the ACM [Volume 18, Number 6, pp. 341–343]

"An Algorithm for Differential File Comparison"
by James W. Hunt and M. Douglas McIlroy, June 1976
Computing Science Technical Report, Bell Laboratories 41

"A Fast Algorithm for Computing Longest Common Subsequences"
by James W. Hunt and Thomas G. Szymanski, published May 1977
Communications of the ACM [Volume 20, Number 5, pp. 350-353]

"A new flexible algorithm for the longest common subsequence problem"
by Claus Rick, published 1995, Proceedings, 6th Annual Symposium on
Combinatorial Pattern Matching [Lecture Notes in Computer Science,
Springer Verlag, Volume 937, pp. 340-351]

"A New Practical Linear Space Algorithm for the Longest Common
Subsequence Problem" by Heiko Goeman and Michael Clausen,
published 2002, Kybernetika [volume 38, Issue 1, pp. 45-66]

'''Hunt and Szymanski algorithm'''

```#include <stdint.h>
#include <string>
#include <memory>                       // for shared_ptr<>
#include <iostream>
#include <deque>
#include <map>
#include <algorithm>                    // for lower_bound()
#include <iterator>                     // for next() and prev()

using namespace std;

class LCS {
protected:
// This linked list class is used to trace the LCS candidates
class Pair {
public:
uint32_t index1;
uint32_t index2;
shared_ptr<Pair> next;

Pair(uint32_t index1, uint32_t index2, shared_ptr<Pair> next = nullptr)
: index1(index1), index2(index2), next(next) {
}

static shared_ptr<Pair> Reverse(const shared_ptr<Pair> pairs) {
for (auto next = pairs; next != nullptr; next = next->next)
}
};

typedef deque<shared_ptr<Pair>> PAIRS;
typedef deque<uint32_t> THRESHOLD;
typedef deque<uint32_t> INDEXES;
typedef map<char, INDEXES> CHAR2INDEXES;
typedef deque<INDEXES*> MATCHES;

// return the LCS as a linked list of matched index pairs
uint32_t Pairs(MATCHES& matches, shared_ptr<Pair> *pairs) {
auto trace = pairs != nullptr;
PAIRS traces;
THRESHOLD threshold;

//
//[Assert]After each index1 iteration threshold[index3] is the least index2
// such that the LCS of s1[0:index1] and s2[0:index2] has length index3 + 1
//
uint32_t index1 = 0;
for (const auto& it1 : matches) {
if (!it1->empty()) {
auto dq2 = *it1;
auto limit = threshold.end();
for (auto it2 = dq2.rbegin(); it2 != dq2.rend(); it2++) {
// Each of the index1, index2 pairs considered here correspond to a match
auto index2 = *it2;

//
// Note: The reverse iterator it2 visits index2 values in descending order,
// allowing thresholds to be updated in-place.  std::lower_bound() is used
// to perform a binary search.
//
limit = lower_bound(threshold.begin(), limit, index2);
auto index3 = distance(threshold.begin(), limit);

//
// Look ahead to the next index2 value to optimize space used in the Hunt
// and Szymanski algorithm.  If the next index2 is also an improvement on
// the value currently held in threshold[index3], a new Pair will only be
// superseded on the next index2 iteration.
//
// Depending on match redundancy, the number of Pair constructions may be
// divided by factors ranging from 2 up to 10 or more.
//
auto skip = next(it2) != dq2.rend() &&
(limit == threshold.begin() || *prev(limit) < *next(it2));
if (skip) continue;

if (limit == threshold.end()) {
// insert case
threshold.push_back(index2);
// Refresh limit iterator:
limit = prev(threshold.end());
if (trace) {
auto prefix = index3 > 0 ? traces[index3 - 1] : nullptr;
auto last = make_shared<Pair>(index1, index2, prefix);
traces.push_back(last);
}
}
else if (index2 < *limit) {
// replacement case
*limit = index2;
if (trace) {
auto prefix = index3 > 0 ? traces[index3 - 1] : nullptr;
auto last = make_shared<Pair>(index1, index2, prefix);
traces[index3] = last;
}
}
}                                 // next index2
}

index1++;
}                                     // next index1

if (trace) {
auto last = traces.size() > 0 ? traces.back() : nullptr;
// Reverse longest back-trace
*pairs = Pair::Reverse(last);
}

auto length = threshold.size();
return length;
}

//
// Match() avoids incurring m*n comparisons by using the associative
// memory implemented by CHAR2INDEXES to achieve O(m+n) performance,
// where m and n are the input lengths.
//
// The lookup time can be assumed constant in the case of characters.
// The symbol space is larger in the case of records; but the lookup
// time will be O(log(m+n)), at most.
//
void Match(CHAR2INDEXES& indexes, MATCHES& matches,
const string& s1, const string& s2) {
uint32_t index = 0;
for (const auto& it : s2)
indexes[it].push_back(index++);

for (const auto& it : s1) {
auto& dq2 = indexes[it];
matches.push_back(&dq2);
}
}

string Select(shared_ptr<Pair> pairs, uint32_t length,
bool right, const string& s1, const string& s2) {
string buffer;
buffer.reserve(length);
for (auto next = pairs; next != nullptr; next = next->next) {
auto c = right ? s2[next->index2] : s1[next->index1];
buffer.push_back(c);
}
return buffer;
}

public:
string Correspondence(const string& s1, const string& s2) {
CHAR2INDEXES indexes;
MATCHES matches;                    // holds references into indexes
Match(indexes, matches, s1, s2);
shared_ptr<Pair> pairs;             // obtain the LCS as index pairs
auto length = Pairs(matches, &pairs);
return Select(pairs, length, false, s1, s2);
}
};
```

'''Example''':

```    LCS lcs;
auto s = lcs.Correspondence(s1, s2);
cout << s << endl;
```

## C#

### With recursion

```using System;

namespace LCS
{
class Program
{
static void Main(string[] args)
{
string word1 = "thisisatest";
string word2 = "testing123testing";

Console.WriteLine(lcsBack(word1, word2));
}

public static string lcsBack(string a, string b)
{
string aSub = a.Substring(0, (a.Length - 1 < 0) ? 0 : a.Length - 1);
string bSub = b.Substring(0, (b.Length - 1 < 0) ? 0 : b.Length - 1);

if (a.Length == 0 || b.Length == 0)
return "";
else if (a[a.Length - 1] == b[b.Length - 1])
return lcsBack(aSub, bSub) + a[a.Length - 1];
else
{
string x = lcsBack(a, bSub);
string y = lcsBack(aSub, b);
return (x.Length > y.Length) ? x : y;
}
}
}
}
```

## Clojure

Based on algorithm from Wikipedia.

```(defn longest [xs ys] (if (> (count xs) (count ys)) xs ys))

(def lcs
(memoize
(fn [[x & xs] [y & ys]]
(cond
(or (= x nil) (= y nil)) nil
(= x y) (cons x (lcs xs ys))
:else (longest (lcs (cons x xs) ys)
(lcs xs (cons y ys)))))))
```

## CoffeeScript

```
lcs = (s1, s2) ->
len1 = s1.length
len2 = s2.length

# Create a virtual matrix that is (len1 + 1) by (len2 + 1),
# where m[i][j] is the longest common string using only
# the first i chars of s1 and first j chars of s2.  The
# matrix is virtual, because we only keep the last two rows
# in memory.
prior_row = ('' for i in [0..len2])

for i in [0...len1]
row = ['']
for j in [0...len2]
if s1[i] == s2[j]
row.push prior_row[j] + s1[i]
else
subs1 = row[j]
subs2 = prior_row[j+1]
if subs1.length > subs2.length
row.push subs1
else
row.push subs2
prior_row = row

row[len2]

s1 = "thisisatest"
s2 = "testing123testing"
console.log lcs(s1, s2)
```

## Common Lisp

Here's a memoizing/dynamic-programming solution that uses an n × m array where n and m are the lengths of the input arrays. The first return value is a sequence (of the same type as array1) which is the longest common subsequence. The second return value is the length of the longest common subsequence.

```(defun longest-common-subsequence (array1 array2)
(let* ((l1 (length array1))
(l2 (length array2))
(results (make-array (list l1 l2) :initial-element nil)))
(declare (dynamic-extent results))
(labels ((lcs (start1 start2)
;; if either sequence is empty, return (() 0)
(if (or (eql start1 l1) (eql start2 l2)) (list '() 0)
;; otherwise, return any memoized value
(let ((result (aref results start1 start2)))
(if (not (null result)) result
;; otherwise, compute and store a value
(setf (aref results start1 start2)
(if (eql (aref array1 start1) (aref array2 start2))
;; move forward in both sequences
(destructuring-bind (seq len)
(lcs (1+ start1) (1+ start2))
(list (cons (aref array1 start1) seq) (1+ len)))
;; otherwise, move ahead in each separately,
;; and return the better result.
(let ((a (lcs (1+ start1) start2))
(b (lcs start1 (1+ start2))))
(if (> (second a) (second b))
a
b)))))))))
(destructuring-bind (seq len) (lcs 0 0)
(values (coerce seq (type-of array1)) len)))))
```

For example,

```(longest-common-subsequence "123456" "1a2b3c")
```

produces the two values

```"123"
3
```

### An alternative adopted from Clojure

Here is another version with its own memoization macro:

```(defmacro mem-defun (name args body)
(let ((hash-name (gensym)))
`(let ((,hash-name (make-hash-table :test 'equal)))
(defun ,name ,args
(or (gethash (list ,@args) ,hash-name)
(setf (gethash (list ,@args) ,hash-name)
,body))))))

(mem-defun lcs (xs ys)
(labels ((longer (a b) (if (> (length a) (length b)) a b)))
(cond ((or (null xs) (null ys)) nil)
((equal (car xs) (car ys)) (cons (car xs) (lcs (cdr xs) (cdr ys))))
(t (longer (lcs (cdr xs) ys)
(lcs xs (cdr ys)))))))
```

When we test it, we get:

```(coerce (lcs (coerce "thisisatest" 'list) (coerce "testing123testing" 'list)) 'string))))

"tsitest"
```

## D

Both versions don't work correctly with Unicode text.

### Recursive version

```import std.stdio, std.array;

T[] lcs(T)(in T[] a, in T[] b) pure nothrow @safe {
if (a.empty || b.empty) return null;
if (a[0] == b[0])
return a[0] ~ lcs(a[1 .. \$], b[1 .. \$]);
const longest = (T[] x, T[] y) => x.length > y.length ? x : y;
return longest(lcs(a, b[1 .. \$]), lcs(a[1 .. \$], b));
}

void main() {
lcs("thisisatest", "testing123testing").writeln;
}
```

{{out}}

```tsitest
```

### Faster dynamic programming version

The output is the same.

```import std.stdio, std.algorithm, std.traits;

T[] lcs(T)(in T[] a, in T[] b) pure /*nothrow*/ {
auto L = new uint[][](a.length + 1, b.length + 1);

foreach (immutable i; 0 .. a.length)
foreach (immutable j; 0 .. b.length)
L[i + 1][j + 1] = (a[i] == b[j]) ? (1 + L[i][j]) :
max(L[i + 1][j], L[i][j + 1]);

Unqual!T[] result;
for (auto i = a.length, j = b.length; i > 0 && j > 0; ) {
if (a[i - 1] == b[j - 1]) {
result ~= a[i - 1];
i--;
j--;
} else
if (L[i][j - 1] < L[i - 1][j])
i--;
else
j--;
}

result.reverse(); // Not nothrow.
return result;
}

void main() {
lcs("thisisatest", "testing123testing").writeln;
}
```

### Hirschberg algorithm version

See: http://en.wikipedia.org/wiki/Hirschberg_algorithm

This is currently a little slower than the classic dynamic programming version, but it uses a linear amount of memory, so it's usable for much larger inputs. To speed up this code on dmd remove the memory allocations from lensLCS, and do not use the retro range (replace it with foreach_reverse). The output is the same.

```import std.stdio, std.algorithm, std.range, std.array, std.string, std.typecons;

uint[] lensLCS(R)(R xs, R ys) pure nothrow @safe {
auto prev = new typeof(return)(1 + ys.length);
auto curr = new typeof(return)(1 + ys.length);

foreach (immutable x; xs) {
swap(curr, prev);
size_t i = 0;
foreach (immutable y; ys) {
curr[i + 1] = (x == y) ? prev[i] + 1 : max(curr[i], prev[i + 1]);
i++;
}
}

return curr;
}

void calculateLCS(T)(in T[] xs, in T[] ys, bool[] xs_in_lcs,
in size_t idx=0) pure nothrow @safe {
immutable nx = xs.length;
immutable ny = ys.length;

if (nx == 0)
return;

if (nx == 1) {
if (ys.canFind(xs[0]))
xs_in_lcs[idx] = true;
} else {
immutable mid = nx / 2;
const xb = xs[0.. mid];
const xe = xs[mid .. \$];
immutable ll_b = lensLCS(xb, ys);

const ll_e = lensLCS(xe.retro, ys.retro); // retro is slow with dmd.

//immutable k = iota(ny + 1)
//              .reduce!(max!(j => ll_b[j] + ll_e[ny - j]));
immutable k = iota(ny + 1)
.minPos!((i, j) => tuple(ll_b[i] + ll_e[ny - i]) >
tuple(ll_b[j] + ll_e[ny - j]))[0];

calculateLCS(xb, ys[0 .. k], xs_in_lcs, idx);
calculateLCS(xe, ys[k .. \$], xs_in_lcs, idx + mid);
}
}

const(T)[] lcs(T)(in T[] xs, in T[] ys) pure /*nothrow*/ @safe {
auto xs_in_lcs = new bool[xs.length];
calculateLCS(xs, ys, xs_in_lcs);
return zip(xs, xs_in_lcs).filter!q{ a[1] }.map!q{ a[0] }.array; // Not nothrow.
}

string lcsString(in string s1, in string s2) pure /*nothrow*/ @safe {
return lcs(s1.representation, s2.representation).assumeUTF;
}

void main() {
lcsString("thisisatest", "testing123testing").writeln;
}
```

## Dart

```import 'dart:math';

String lcsRecursion(String a, String b) {
int aLen = a.length;
int bLen = b.length;

if (aLen == 0 || bLen == 0) {
return "";
} else if (a[aLen-1] == b[bLen-1]) {
return lcsRecursion(a.substring(0,aLen-1),b.substring(0,bLen-1)) + a[aLen-1];
} else {
var x = lcsRecursion(a, b.substring(0,bLen-1));
var y = lcsRecursion(a.substring(0,aLen-1), b);
return (x.length > y.length) ? x : y;
}
}

String lcsDynamic(String a, String b) {
var lengths = new List<List<int>>.generate(a.length + 1,
(_) => new List.filled(b.length+1, 0), growable: false);

// row 0 and column 0 are initialized to 0 already
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < b.length; j++) {
if (a[i] == b[j]) {
lengths[i+1][j+1] = lengths[i][j] + 1;
} else {
lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1]);
}
}
}

// read the substring out from the matrix
StringBuffer reversedLcsBuffer = new StringBuffer();
for (int x = a.length, y = b.length; x != 0 && y != 0;) {
if (lengths[x][y] == lengths[x-1][y]) {
x--;
} else if (lengths[x][y] == lengths[x][y-1]) {
y--;
} else {
assert(a[x-1] == b[y-1]);
reversedLcsBuffer.write(a[x-1]);
x--;
y--;
}
}

// reverse String
var reversedLCS = reversedLcsBuffer.toString();
var lcsBuffer = new StringBuffer();
for(var i = reversedLCS.length - 1; i>=0; i--) {
lcsBuffer.write(reversedLCS[i]);
}
return lcsBuffer.toString();
}

void main() {
print("lcsDynamic('1234', '1224533324') =  \${lcsDynamic('1234', '1224533324')}");
print("lcsDynamic('thisisatest', 'testing123testing') = \${lcsDynamic('thisisatest', 'testing123testing')}");
print("lcsDynamic('', 'x') = \${lcsDynamic('', 'x')}");
print("lcsDynamic('x', 'x') = \${lcsDynamic('x', 'x')}");
print('');
print("lcsRecursion('1234', '1224533324') = \${lcsRecursion('1234', '1224533324')}");
print("lcsRecursion('thisisatest', 'testing123testing') = \${lcsRecursion('thisisatest', 'testing123testing')}");
print("lcsRecursion('', 'x') = \${lcsRecursion('', 'x')}");
print("lcsRecursion('x', 'x') = \${lcsRecursion('x', 'x')}");
}

```

{{out}}

```lcsDynamic('1234', '1224533324') = 1234
lcsDynamic('thisisatest', 'testing123testing') = tsitest
lcsDynamic('', 'x') =
lcsDynamic('x', 'x') = x

lcsRecursion('1234', '1224533324') = 1234
lcsRecursion('thisisatest', 'testing123testing') = tsitest
lcsRecursion('', 'x') =
lcsRecursion('x', 'x') = x
```

## Egison

```
(define \$common-seqs
(lambda [\$xs \$ys]
(match-all [xs ys] [(list char) (list char)]
[[(loop \$i [1 \$n] <join _ <cons \$c_i ...>> _)
(loop \$i [1 ,n] <join _ <cons ,c_i ...>> _)]
(map (lambda [\$i] c_i) (between 1 n))])))

(define \$lcs (compose common-seqs rac))

```

'''Output:'''

```
> (lcs "thisisatest" "testing123testing"))
"tsitest"

```

## Elixir

{{works with|Elixir|1.3}}

### Simple recursion

This solution is Brute force. It is slow {{trans|Ruby}}

```defmodule LCS do
def lcs(a, b) do
lcs(to_charlist(a), to_charlist(b), []) |> to_string
end

defp lcs([h|at], [h|bt], res), do: lcs(at, bt, [h|res])
defp lcs([_|at]=a, [_|bt]=b, res) do
Enum.max_by([lcs(a, bt, res), lcs(at, b, res)], &length/1)
end
defp lcs(_, _, res), do: res |> Enum.reverse
end

IO.puts LCS.lcs("thisisatest", "testing123testing")
IO.puts LCS.lcs('1234','1224533324')
```

### Dynamic Programming

{{trans|Erlang}}

```defmodule LCS do
def lcs_length(s,t), do: lcs_length(s,t,Map.new) |> elem(0)

defp lcs_length([],t,cache), do: {0,Map.put(cache,{[],t},0)}
defp lcs_length(s,[],cache), do: {0,Map.put(cache,{s,[]},0)}
defp lcs_length([h|st]=s,[h|tt]=t,cache) do
{l,c} = lcs_length(st,tt,cache)
{l+1,Map.put(c,{s,t},l+1)}
end
defp lcs_length([_sh|st]=s,[_th|tt]=t,cache) do
if Map.has_key?(cache,{s,t}) do
{Map.get(cache,{s,t}),cache}
else
{l1,c1} = lcs_length(s,tt,cache)
{l2,c2} = lcs_length(st,t,c1)
l = max(l1,l2)
{l,Map.put(c2,{s,t},l)}
end
end

def lcs(s,t) do
{s,t} = {to_charlist(s),to_charlist(t)}
{_,c} = lcs_length(s,t,Map.new)
lcs(s,t,c,[]) |> to_string
end

defp lcs([],_,_,acc), do: Enum.reverse(acc)
defp lcs(_,[],_,acc), do: Enum.reverse(acc)
defp lcs([h|st],[h|tt],cache,acc), do: lcs(st,tt,cache,[h|acc])
defp lcs([_sh|st]=s,[_th|tt]=t,cache,acc) do
if Map.get(cache,{s,tt}) > Map.get(cache,{st,t}) do
lcs(s,tt,cache,acc)
else
lcs(st,t,cache,acc)
end
end
end

IO.puts LCS.lcs("thisisatest","testing123testing")
IO.puts LCS.lcs("1234","1224533324")
```

{{out}}

```
tsitest
1234

```

Referring to LCS [[Shortest common supersequence#Elixir|here]].

## Erlang

This implementation also includes the ability to calculate the length of the longest common subsequence. In calculating that length, we generate a cache which can be traversed to generate the longest common subsequence.

```
module(lcs).
-compile(export_all).

lcs_length(S,T) ->
{L,_C} = lcs_length(S,T,dict:new()),
L.

lcs_length([]=S,T,Cache) ->
{0,dict:store({S,T},0,Cache)};
lcs_length(S,[]=T,Cache) ->
{0,dict:store({S,T},0,Cache)};
lcs_length([H|ST]=S,[H|TT]=T,Cache) ->
{L,C} = lcs_length(ST,TT,Cache),
{L+1,dict:store({S,T},L+1,C)};
lcs_length([_SH|ST]=S,[_TH|TT]=T,Cache) ->
case dict:is_key({S,T},Cache) of
true -> {dict:fetch({S,T},Cache),Cache};
false ->
{L1,C1} = lcs_length(S,TT,Cache),
{L2,C2} = lcs_length(ST,T,C1),
L = lists:max([L1,L2]),
{L,dict:store({S,T},L,C2)}
end.

lcs(S,T) ->
{_,C} = lcs_length(S,T,dict:new()),
lcs(S,T,C,[]).

lcs([],_,_,Acc) ->
lists:reverse(Acc);
lcs(_,[],_,Acc) ->
lists:reverse(Acc);
lcs([H|ST],[H|TT],Cache,Acc) ->
lcs(ST,TT,Cache,[H|Acc]);
lcs([_SH|ST]=S,[_TH|TT]=T,Cache,Acc) ->
case dict:fetch({S,TT},Cache) > dict:fetch({ST,T},Cache) of
true ->
lcs(S,TT,Cache, Acc);
false ->
lcs(ST,T,Cache,Acc)
end.

```

'''Output:'''

```
77> lcs:lcs("thisisatest","testing123testing").
"tsitest"
78> lcs:lcs("1234","1224533324").
"1234"

```

We can also use the process dictionary to memoize the recursive implementation:

```
lcs(Xs0, Ys0) ->
CacheKey = {lcs_cache, Xs0, Ys0},
case get(CacheKey)
of  undefined ->
Result =
case {Xs0, Ys0}
of  {[], _} -> []
;   {_, []} -> []
;   {[Same | Xs], [Same | Ys]} ->
[Same | lcs(Xs, Ys)]
;   {[_ | XsRest]=XsAll, [_ | YsRest]=YsAll} ->
A = lcs(XsRest, YsAll),
B = lcs(XsAll , YsRest),
case length(A) > length(B)
of  true  -> A
;   false -> B
end
end,
undefined = put(CacheKey, Result),
Result
;   Result ->
Result
end.

```

Similar to the above, but without using the process dictionary:

```
-module(lcs).

%% API exports
-export([
lcs/2
]).

%%
### ==============================================================

%% API functions
%%
### ==============================================================

lcs(A, B) ->
{LCS, _Cache} = get_lcs(A, B, [], #{}),
lists:reverse(LCS).

%%
### ==============================================================

%% Internal functions
%%
### ===============================================

get_lcs(A, B, Acc, Cache) ->
case maps:find({A, B, Acc}, Cache) of
{ok, LCS} -> {LCS, Cache};
error     ->
{NewLCS, NewCache} = compute_lcs(A, B, Acc, Cache),
{NewLCS, NewCache#{ {A, B, Acc} => NewLCS }}
end.

compute_lcs(A, B, Acc, Cache) when length(A) == 0 orelse length(B) == 0 ->
{Acc, Cache};
compute_lcs([Token |ATail], [Token |BTail], Acc, Cache) ->
get_lcs(ATail, BTail, [Token |Acc], Cache);
compute_lcs([_AToken |ATail]=A, [_BToken |BTail]=B, Acc, Cache) ->
{LCSA, CacheA} = get_lcs(A, BTail, Acc, Cache),
{LCSB, CacheB} = get_lcs(ATail, B, Acc, CacheA),
LCS = case length(LCSA) > length(LCSB) of
true  -> LCSA;
false -> LCSB
end,
{LCS, CacheB}.

```

'''Output:'''

```
48> lcs:lcs("thisisatest", "testing123testing").
"tsitest"

```

## Fortran

{{works with|Fortran|95}} Using the iso_varying_string module which can be found [http://www.fortran.com/iso_varying_string.f95 here] (or equivalent module conforming to the ISO/IEC 1539-2:2000 API or to a subset according to the need of this code: `char`, `len`, `//`, `extract`, `==`, `=`)

```program lcstest
use iso_varying_string
implicit none

type(varying_string) :: s1, s2

s1 = "thisisatest"
s2 = "testing123testing"
print *, char(lcs(s1, s2))

s1 = "1234"
s2 = "1224533324"
print *, char(lcs(s1, s2))

contains

recursive function lcs(a, b) result(l)
type(varying_string) :: l
type(varying_string), intent(in) :: a, b

type(varying_string) :: x, y

l = ""
if ( (len(a) == 0) .or. (len(b) == 0) ) return
if ( extract(a, len(a), len(a)) == extract(b, len(b), len(b)) ) then
l = lcs(extract(a, 1, len(a)-1), extract(b, 1, len(b)-1)) // extract(a, len(a), len(a))
else
x = lcs(a, extract(b, 1, len(b)-1))
y = lcs(extract(a, 1, len(a)-1), b)
if ( len(x) > len(y) ) then
l = x
else
l = y
end if
end if
end function lcs

end program lcstest
```

```open System

let longest xs ys = if List.length xs > List.length ys then xs else ys

let rec lcs a b =
match a, b with
| [], _
| _, []        -> []
| x::xs, y::ys ->
if x = y then
x :: lcs xs ys
else
longest (lcs a ys) (lcs xs b)

[<EntryPoint>]
let main argv =
let split (str:string) = List.init str.Length (fun i -> str.[i])
printfn "%A" (String.Join("",
(lcs (split "thisisatest") (split "testing123testing"))))
0

```

## Factor

```USE: lcs
"thisisatest" "testing123testing" lcs print
```

{{out}}

```
tsitest

```

## Go

{{trans|Java}}

### Recursion

Brute force

```func lcs(a, b string) string {
aLen := len(a)
bLen := len(b)
if aLen == 0 || bLen == 0 {
return ""
} else if a[aLen-1] == b[bLen-1] {
return lcs(a[:aLen-1], b[:bLen-1]) + string(a[aLen-1])
}
x := lcs(a, b[:bLen-1])
y := lcs(a[:aLen-1], b)
if len(x) > len(y) {
return x
}
return y
}
```

### Dynamic Programming

```func lcs(a, b string) string {
arunes := []rune(a)
brunes := []rune(b)
aLen := len(arunes)
bLen := len(brunes)
lengths := make([][]int, aLen+1)
for i := 0; i <= aLen; i++ {
lengths[i] = make([]int, bLen+1)
}
// row 0 and column 0 are initialized to 0 already

for i := 0; i < aLen; i++ {
for j := 0; j < bLen; j++ {
if arunes[i] == brunes[j] {
lengths[i+1][j+1] = lengths[i][j] + 1
} else if lengths[i+1][j] > lengths[i][j+1] {
lengths[i+1][j+1] = lengths[i+1][j]
} else {
lengths[i+1][j+1] = lengths[i][j+1]
}
}
}

// read the substring out from the matrix
s := make([]rune, 0, lengths[aLen][bLen])
for x, y := aLen, bLen; x != 0 && y != 0; {
if lengths[x][y] == lengths[x-1][y] {
x--
} else if lengths[x][y] == lengths[x][y-1] {
y--
} else {
s = append(s, arunes[x-1])
x--
y--
}
}
// reverse string
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
return string(s)
}
```

## Groovy

Recursive solution:

```def lcs(xstr, ystr) {
if (xstr == "" || ystr == "") {
return "";
}

def x = xstr[0];
def y = ystr[0];

def xs = xstr.size() > 1 ? xstr[1..-1] : "";
def ys = ystr.size() > 1 ? ystr[1..-1] : "";

if (x == y) {
return (x + lcs(xs, ys));
}

def lcs1 = lcs(xstr, ys);
def lcs2 = lcs(xs, ystr);

lcs1.size() > lcs2.size() ? lcs1 : lcs2;
}

println(lcs("1234", "1224533324"));
println(lcs("thisisatest", "testing123testing"));
```

{{out}}

```1234
tsitest
```

The [[wp:Longest_common_subsequence#Solution_for_two_sequences|Wikipedia solution]] translates directly into Haskell, with the only difference that equal characters are added in front:

``` length ys then xs else ys

lcs [] _ = []
lcs _ [] = []
lcs (x:xs) (y:ys)
| x == y    = x : lcs xs ys
| otherwise = longest (lcs (x:xs) ys) (lcs xs (y:ys))
```

A Memoized version of the naive algorithm.

```import qualified Data.MemoCombinators as M

lcs = memoize lcsm
where
lcsm [] _ = []
lcsm _ [] = []
lcsm (x:xs) (y:ys)
| x == y    = x : lcs xs ys
| otherwise = maxl (lcs (x:xs) ys) (lcs xs (y:ys))

maxl x y = if length x > length y then x else y
memoize = M.memo2 mString mString
mString = M.list M.char -- Chars, but you can specify any type you need for the memo
```

Memoization (aka dynamic programming) of that uses ''zip'' to make both the index and the character available:

```import Data.Array

lcs xs ys = a!(0,0) where
n = length xs
m = length ys
a = array ((0,0),(n,m)) \$ l1 ++ l2 ++ l3
l1 = [((i,m),[]) | i <- [0..n]]
l2 = [((n,j),[]) | j <- [0..m]]
l3 = [((i,j), f x y i j) | (x,i) <- zip xs [0..], (y,j) <- zip ys [0..]]
f x y i j
| x == y    = x : a!(i+1,j+1)
| otherwise = longest (a!(i,j+1)) (a!(i+1,j))
```

All 3 solutions work of course not only with strings, but also with any other list. Example:

```*Main> lcs "thisisatest" "testing123testing"
"tsitest"
```

The dynamic programming version without using arrays:

```import Data.List

longest xs ys = if length xs > length ys then xs else ys

lcs xs ys = head \$ foldr(\xs -> map head. scanr1 f. zipWith (\x y -> [x,y]) xs) e m where
m = map (\x -> flip (++) [[]] \$ map (\y -> [x | x==y]) ys) xs
e = replicate (length ys) []
f [a,b] [c,d]
| null a = longest b c: [b]
| otherwise = (a++d):[b]
```

Simple and slow solution:

```import Data.Ord
import Data.List

--          longest                        common
lcs xs ys = maximumBy (comparing length) \$ intersect (subsequences xs) (subsequences ys)

main = print \$ lcs "thisisatest" "testing123testing"
```

{{out}}

```"tsitest"
```

=={{header|Icon}} and {{header|Unicon}}== This solution is a modified variant of the recursive solution. The modifications include (a) deleting all characters not common to both strings and (b) stripping off common prefixes and suffixes in a single step.

{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/strings.icn Uses deletec from strings]

```procedure main()
LCSTEST("thisisatest","testing123testing")
LCSTEST("","x")
LCSTEST("x","x")
LCSTEST("beginning-middle-ending","beginning-diddle-dum-ending")
end

procedure LCSTEST(a,b)    #: helper to show inputs and results
write("lcs( ",image(a),", ",image(b)," ) = ",image(res := lcs(a,b)))
return res
end

procedure lcs(a,b)     #: return longest common sub-sequence of characters (modified recursive method)
local i,x,y
local c,nc

if *(a|b) = 0 then return ""                               # done if either string is empty
if a == b then return a                                    # done if equal

if *(a ++ b -- (c := a ** b)) > 0 then {                   # find all characters not in common
a := deletec(a,nc := ~c)                                # .. remove
b := deletec(b,nc)                                      # .. remove
}                                                       # only unequal strings and shared characters beyond

i := 0 ; while a[i+1] == b[i+1] do i +:=1                  # find common prefix ...
if *(x := a[1+:i]) > 0  then                               # if any
return x || lcs(a[i+1:0],b[i+1:0])                      # ... remove and process remainder

i := 0 ; while a[-(i+1)] == b[-(i+1)] do i +:=1            # find common suffix ...
if *(y := a[0-:i]) > 0 then                                # if any
return lcs(a[1:-i],b[1:-i]) || y                        # ... remove and process remainder

return if *(x := lcs(a,b[1:-1])) > *(y := lcs(a[1:-1],b)) then x else y  # divide, discard, and keep longest
end
```

{{out}}

```lcs( "thisisatest", "testing123testing" ) = "tsitest"
lcs( "", "x" ) = ""
lcs( "x", "x" ) = "x"
lcs( "beginning-middle-ending", "beginning-diddle-dum-ending" ) = "beginning-iddle-ending"
```

## J

```lcs=: dyad define
|.x{~ 0{"1 cullOne^:_ (\: +/"1)(\:{."1) 4\$.\$. x =/ y
)

cullOne=: ({~[: <@<@< [: (i. 0:)1,[: *./[: |: 2>/\]) :: ]
```

Here's [[Longest_common_subsequence/J|another approach]]:

```mergeSq=: ;@}:  ~.@, {.@;@{. ,&.> 3 {:: 4&{.
common=: 2 2 <@mergeSq@,;.3^:_ [: (<@#&.> i.@\$) =/
lcs=: [ {~ 0 {"1 ,&\$ #: 0 ({:: (#~ [: (= >./) #@>)) 0 ({:: ,) common
```

Example use (works with either definition of lcs):

```   'thisisatest' lcs 'testing123testing'
tsitest
```

'''Dynamic programming version'''

```longest=: ]`[@.(>&#)
upd=:{:@[,~ ({.@[ ,&.> {:@])`({:@[ longest&.> {.@])@.(0 = #&>@{.@[)
lcs=: 0{:: [: ([: {.&> [: upd&.>/\.<"1@:,.)/ a:,.~a:,~=/{"1 a:,.<"0@[
```

'''Output:'''

```   '1234' lcs '1224533324'
1234

'thisisatest' lcs 'testing123testing'
tsitest
```

'''Recursion'''

```lcs=:;((\$:}.) longest }.@[ \$: ])`({.@[,\$:&}.)@.(=&{.)`((i.0)"_)@.(+.&(0=#))&((e.#[)&>/) ;~
```

## Java

### Recursion

This is not a particularly fast algorithm, but it gets the job done eventually. The speed is a result of many recursive function calls.

```public static String lcs(String a, String b){
int aLen = a.length();
int bLen = b.length();
if(aLen == 0 || bLen == 0){
return "";
}else if(a.charAt(aLen-1) == b.charAt(bLen-1)){
return lcs(a.substring(0,aLen-1),b.substring(0,bLen-1))
+ a.charAt(aLen-1);
}else{
String x = lcs(a, b.substring(0,bLen-1));
String y = lcs(a.substring(0,aLen-1), b);
return (x.length() > y.length()) ? x : y;
}
}
```

### Dynamic Programming

```public static String lcs(String a, String b) {
int[][] lengths = new int[a.length()+1][b.length()+1];

// row 0 and column 0 are initialized to 0 already

for (int i = 0; i < a.length(); i++)
for (int j = 0; j < b.length(); j++)
if (a.charAt(i) == b.charAt(j))
lengths[i+1][j+1] = lengths[i][j] + 1;
else
lengths[i+1][j+1] =
Math.max(lengths[i+1][j], lengths[i][j+1]);

// read the substring out from the matrix
StringBuffer sb = new StringBuffer();
for (int x = a.length(), y = b.length();
x != 0 && y != 0; ) {
if (lengths[x][y] == lengths[x-1][y])
x--;
else if (lengths[x][y] == lengths[x][y-1])
y--;
else {
assert a.charAt(x-1) == b.charAt(y-1);
sb.append(a.charAt(x-1));
x--;
y--;
}
}

return sb.reverse().toString();
}
```

## JavaScript

### Recursion

{{trans|Java}} This is more or less a translation of the recursive Java version above.

```function lcs(a, b) {
var aSub = a.substr(0, a.length - 1);
var bSub = b.substr(0, b.length - 1);

if (a.length === 0 || b.length === 0) {
return '';
} else if (a.charAt(a.length - 1) === b.charAt(b.length - 1)) {
return lcs(aSub, bSub) + a.charAt(a.length - 1);
} else {
var x = lcs(a, bSub);
var y = lcs(aSub, b);
return (x.length > y.length) ? x : y;
}
}
```

ES6 recursive implementation

```
var longest = (xs, ys) => (xs.length > ys.length) ? xs : ys;

var lcs = (xx, yy) => {
if (!xx.length || !yy.length) { return ''; }

var x = xx[0],
y = yy[0];
xs = xx.slice(1);
ys = yy.slice(1);

return (x === y) ? lcs(xs, ys) :
longest(lcs(xx, ys), lcs(xs, yy));
};
```

### Dynamic Programming

This version runs in O(mn) time and consumes O(mn) space. Factoring out loop edge cases could get a small constant time improvement, and it's fairly trivial to edit the final loop to produce a full diff in addition to the lcs.

```function lcs(x,y){
var s,i,j,m,n,
lcs=[],row=[],c=[],
left,diag,latch;
//make sure shorter string is the column string
if(m<n){s=x;x=y;y=s;}
m = x.length;
n = y.length;
//build the c-table
for(j=0;j<n;row[j++]=0);
for(i=0;i<m;i++){
c[i] = row = row.slice();
for(diag=0,j=0;j<n;j++,diag=latch){
latch=row[j];
if(x[i] == y[j]){row[j] = diag+1;}
else{
left = row[j-1]||0;
if(left>row[j]){row[j] = left;}
}
}
}
i--,j--;
//row[j] now contains the length of the lcs
//recover the lcs from the table
while(i>-1&&j>-1){
switch(c[i][j]){
default: j--;
lcs.unshift(x[i]);
case (i&&c[i-1][j]): i--;
continue;
case (j&&c[i][j-1]): j--;
}
}
return lcs.join('');
}
```

'''BUG note: In line 6, m and n are not yet initialized, and so x and y are never swapped.''' '''Swapping is useless here, and becomes wrong when extending the algorithm to produce a diff.'''

The final loop can be modified to concatenate maximal common substrings rather than individual characters:

```	var t=i;
while(i>-1&&j>-1){
switch(c[i][j]){
default:i--,j--;
continue;
case (i&&c[i-1][j]):
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}
t=--i;
continue;
case (j&&c[i][j-1]): j--;
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}
t=i;
}
}
if(t!==i){lcs.unshift(x.substring(i+1,t+1));}
```

### Greedy Algorithm

This is an heuristic algorithm that won't always return the correct answer, but is significantly faster and less memory intensive than the dynamic programming version, in exchange for giving up the ability to re-use the table to find alternate solutions and greater complexity in generating diffs. Note that this implementation uses a binary buffer for additional efficiency gains, but it's simple to transform to use string or array concatenation;

```function lcs_greedy(x,y){
var p1, i, idx,
symbols = {},
r = 0,
p = 0,
l = 0,
m = x.length,
n = y.length,
s = new Buffer((m < n) ? n : m);

p1 = popsym(0);

for (i = 0; i < m; i++) {
p = (r === p) ? p1 : popsym(i);
p1 = popsym(i + 1);
if (p > p1) {
i += 1;
idx = p1;
} else {
idx = p;
}

if (idx === n) {
p = popsym(i);
} else {
r = idx;
s[l] = x.charCodeAt(i);
l += 1;
}
}
return s.toString('utf8', 0, l);

function popsym(index) {
var s = x[index],
pos = symbols[s] + 1;

pos = y.indexOf(s, ((pos > r) ? pos : r));
if (pos === -1) { pos = n; }
symbols[s] = pos;
return pos;
}
}
```

Note that it won't return the correct answer for all inputs. For example:

```lcs_greedy('bcaaaade', 'deaaaabc'); // 'bc' instead of 'aaaa'
```

## jq

Naive recursive version:

```def lcs(xstr; ystr):
if (xstr == "" or ystr == "") then ""
else
xstr[0:1] as \$x
|  xstr[1:] as \$xs
|  ystr[1:] as \$ys
| if (\$x == ystr[0:1]) then (\$x + lcs(\$xs; \$ys))
else
lcs(xstr; \$ys) as \$one
| lcs(\$xs; ystr) as \$two
| if (\$one|length) > (\$two|length) then \$one else \$two end
end
end;
```

Example:

```lcs("1234"; "1224533324"),
lcs("thisisatest"; "testing123testing")
```

Output:

```
# jq -n -f lcs-recursive.jq
"1234"
"tsitest"
```

## Julia

{{works with|Julia|0.6}}

```longest(a::String, b::String) = length(a) ≥ length(b) ? a : b

"""
julia> lcsrecursive("thisisatest", "testing123testing")
"tsitest"
"""
# Recursive
function lcsrecursive(xstr::String, ystr::String)
if length(xstr) == 0 || length(ystr) == 0
return ""
end

x, xs, y, ys = xstr[1], xstr[2:end], ystr[1], ystr[2:end]
if x == y
return string(x, lcsrecursive(xs, ys))
else
return longest(lcsrecursive(xstr, ys), lcsrecursive(xs, ystr))
end
end

# Dynamic
function lcsdynamic(a::String, b::String)
lengths = zeros(Int, length(a) + 1, length(b) + 1)

# row 0 and column 0 are initialized to 0 already
for (i, x) in enumerate(a), (j, y) in enumerate(b)
if x == y
lengths[i+1, j+1] = lengths[i, j] + 1
else
lengths[i+1, j+1] = max(lengths[i+1, j], lengths[i, j+1])
end
end

# read the substring out from the matrix
result = ""
x, y = length(a) + 1, length(b) + 1
while x > 1 && y > 1
if lengths[x, y] == lengths[x-1, y]
x -= 1
elseif lengths[x, y] == lengths[x, y-1]
y -= 1
else
@assert a[x-1] == b[y-1]
result = string(a[x-1], result)
x -= 1
y -= 1
end
end

return result
end

@show lcsrecursive("thisisatest", "testing123testing")
@time lcsrecursive("thisisatest", "testing123testing")
@show lcsdynamic("thisisatest", "testing123testing")
@time lcsdynamic("thisisatest", "testing123testing")
```

{{out}}

```lcsrecursive("thisisatest", "testing123testing") = "tsitest"
0.038153 seconds (537.77 k allocations: 16.415 MiB)
lcsdynamic("thisisatest", "testing123testing") = "tsitest"
0.000004 seconds (12 allocations: 2.141 KiB)
```

## Kotlin

```// version 1.1.2

fun lcs(x: String, y: String): String {
if (x.length == 0 || y.length == 0) return ""
val x1 = x.dropLast(1)
val y1 = y.dropLast(1)
if (x.last() == y.last()) return lcs(x1, y1) + x.last()
val x2 = lcs(x, y1)
val y2 = lcs(x1, y)
return if (x2.length > y2.length) x2 else y2
}

fun main(args: Array<String>) {
val x = "thisisatest"
val y = "testing123testing"
println(lcs(x, y))
}
```

{{out}}

```
tsitest

```

## Liberty BASIC

```
'variation of BASIC example
w\$="aebdef"
z\$="cacbc"
print lcs\$(w\$,z\$)

'output:
'ab

wait

FUNCTION lcs\$(a\$, b\$)
IF LEN(a\$) = 0 OR LEN(b\$) = 0 THEN
lcs\$ = ""
exit function
end if

IF RIGHT\$(a\$, 1) = RIGHT\$(b\$, 1) THEN
lcs\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), LEFT\$(b\$, LEN(b\$) - 1)) + RIGHT\$(a\$, 1)
exit function
ELSE
x\$ = lcs\$(a\$, LEFT\$(b\$, LEN(b\$) - 1))
y\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), b\$)
IF LEN(x\$) > LEN(y\$) THEN
lcs\$ = x\$
exit function
ELSE
lcs\$ = y\$
exit function
END IF
END IF
END FUNCTION

```

This implementation works on both words and lists.

```to longest :s :t
output ifelse greater? count :s count :t [:s] [:t]
end
to lcs :s :t
if empty? :s [output :s]
if empty? :t [output :t]
if equal? first :s first :t [output combine  first :s  lcs bf :s bf :t]
output longest lcs :s bf :t  lcs bf :s :t
end
```

## Lua

```function LCS( a, b )
if #a == 0 or #b == 0 then
return ""
elseif string.sub( a, -1, -1 ) == string.sub( b, -1, -1 ) then
return LCS( string.sub( a, 1, -2 ), string.sub( b, 1, -2 ) ) .. string.sub( a, -1, -1 )
else
local a_sub = LCS( a, string.sub( b, 1, -2 ) )
local b_sub = LCS( string.sub( a, 1, -2 ), b )

if #a_sub > #b_sub then
return a_sub
else
return b_sub
end
end
end

print( LCS( "thisisatest", "testing123testing" ) )
```

## M4

```define(`set2d',`define(`\$1[\$2][\$3]',`\$4')')
define(`get2d',`defn(\$1[\$2][\$3])')
define(`tryboth',
`pushdef(`x',lcs(`\$1',substr(`\$2',1),`\$1 \$2'))`'pushdef(`y',
lcs(substr(`\$1',1),`\$2',`\$1 \$2'))`'ifelse(eval(len(x)>len(y)),1,
`x',`y')`'popdef(`x')`'popdef(`y')')
define(`checkfirst',
`ifelse(substr(`\$1',0,1),substr(`\$2',0,1),
`substr(`\$1',0,1)`'lcs(substr(`\$1',1),substr(`\$2',1))',
`tryboth(`\$1',`\$2')')')
define(`lcs',
`ifelse(get2d(`c',`\$1',`\$2'),`',
`pushdef(`a',ifelse(
`\$1',`',`',
`\$2',`',`',
`checkfirst(`\$1',`\$2')'))`'a`'set2d(`c',`\$1',`\$2',a)`'popdef(`a')',
`get2d(`c',`\$1',`\$2')')')

lcs(`1234',`1224533324')

lcs(`thisisatest',`testing123testing')
```

Note: the caching (set2d/get2d) obscures the code even more than usual, but is necessary in order to get the second test to run in a reasonable amount of time.

## Maple

```
> StringTools:-LongestCommonSubSequence( "thisisatest", "testing123testing" );
"tsitest"

```

## Mathematica

A built-in function can do this for us:

```a = "thisisatest";
b = "testing123testing";
LongestCommonSequence[a, b]
```

gives:

```
Note that Mathematica also has a built-in function called LongestCommonSubsequence[a,b]:

''finds the longest contiguous subsequence of elements common to the strings or lists a and b.''

which would give "test" as the result for LongestCommonSubsequence[a, b].

The description for LongestCommonSequence[a,b] is:

''finds the longest sequence of contiguous or disjoint elements common to the strings or lists a and b.''

I added this note because the name of this article suggests LongestCommonSubsequence does the job, however LongestCommonSubsequence performs the puzzle-description.

## Nim

### Recursion

{{trans|Python}}

```nim
proc lcs(x, y): string =
if x == "" or y == "":
return ""

if x[0] == y[0]:
return x[0] & lcs(x[1..x.high], y[1..y.high])

let a = lcs(x, y[1..y.high])
let b = lcs(x[1..x.high], y)
result = if a.len > b.len: a else: b

echo lcs("1234", "1224533324")
echo lcs("thisisatest", "testing123testing")
```

### Dynamic Programming

{{trans|Python}}

```proc lcs(a, b): string =
var ls = newSeq[seq[int]] a.len+1
for i in 0 .. a.len:
ls[i].newSeq b.len+1

for i, x in a:
for j, y in b:
if x == y:
ls[i+1][j+1] = ls[i][j] + 1
else:
ls[i+1][j+1] = max(ls[i+1][j], ls[i][j+1])

result = ""
var x = a.len
var y = b.len
while x > 0 and y > 0:
if ls[x][y] == ls[x-1][y]:
dec x
elif ls[x][y] == ls[x][y-1]:
dec y
else:
assert a[x-1] == b[y-1]
result = a[x-1] & result
dec x
dec y

echo lcs("1234", "1224533324")
echo lcs("thisisatest", "testing123testing")
```

## OCaml

### Recursion

``` List.length ys then xs else ys

let rec lcs a b = match a, b with
[], _
| _, []        -> []
| x::xs, y::ys ->
if x = y then
x :: lcs xs ys
else
longest (lcs a ys) (lcs xs b)
```

### Memoized recursion

```
let lcs xs ys =
let cache = Hashtbl.create 16 in
let rec lcs xs ys =
try Hashtbl.find cache (xs, ys) with
| Not_found ->
let result =
match xs, ys with
| [], _ -> []
| _, [] -> []
| x :: xs, y :: ys when x = y ->
x :: lcs xs ys
| _ :: xs_rest, _ :: ys_rest ->
let a = lcs xs_rest ys in
let b = lcs xs      ys_rest in
if (List.length a) > (List.length b) then a else b
in
result
in
lcs xs ys
```

### Dynamic programming

```let lcs xs' ys' =
let xs = Array.of_list xs'
and ys = Array.of_list ys' in
let n = Array.length xs
and m = Array.length ys in
let a = Array.make_matrix (n+1) (m+1) [] in
for i = n-1 downto 0 do
for j = m-1 downto 0 do
a.(i).(j) <- if xs.(i) = ys.(j) then
xs.(i) :: a.(i+1).(j+1)
else
longest a.(i).(j+1) a.(i+1).(j)
done
done;
a.(0).(0)
```

Because both solutions only work with lists, here are some functions to convert to and from strings:

```let list_of_string str =
let result = ref [] in
String.iter (fun x -> result := x :: !result)
str;
List.rev !result

let string_of_list lst =
let result = String.create (List.length lst) in
ignore (List.fold_left (fun i x -> result.[i] <- x; i+1) 0 lst);
result
```

Both solutions work. Example:

```
# string_of_list (lcs (list_of_string "thisisatest")
(list_of_string "testing123testing"));;
- : string = "tsitest"

```

## Oz

Recursive solution:

```declare
fun {LCS Xs Ys}
case [Xs Ys]
of [nil _]                   then nil
[] [_ nil]                   then nil
[] [X|Xr  Y|Yr] andthen X==Y then X|{LCS Xr Yr}
[] [_|Xr  _|Yr]              then {Longest {LCS Xs Yr} {LCS Xr Ys}}
end
end

fun {Longest Xs Ys}
if {Length Xs} > {Length Ys} then Xs else Ys end
end
in
{System.showInfo {LCS "thisisatest" "testing123testing"}}
```

## Pascal

{{trans|Fortran}}

```Program LongestCommonSubsequence(output);

function lcs(a, b: string): string;
var
x, y: string;
lenga, lengb: integer;
begin
lenga := length(a);
lengb := length(b);
lcs := '';
if (lenga >  0) and (lengb >  0) then
if a[lenga] =  b[lengb] then
lcs := lcs(copy(a, 1, lenga-1), copy(b, 1, lengb-1)) + a[lenga]
else
begin
x := lcs(a, copy(b, 1, lengb-1));
y := lcs(copy(a, 1, lenga-1), b);
if length(x) > length(y) then
lcs := x
else
lcs := y;
end;
end;

var
s1, s2: string;
begin
s1 := 'thisisatest';
s2 := 'testing123testing';
writeln (lcs(s1, s2));
s1 := '1234';
s2 := '1224533324';
writeln (lcs(s1, s2));
end.
```

{{out}}

```:> ./LongestCommonSequence
tsitest
1234

```

## Perl

```sub lcs {
my (\$a, \$b) = @_;
if (!length(\$a) || !length(\$b)) {
return "";
}
if (substr(\$a, 0, 1) eq substr(\$b, 0, 1)) {
return substr(\$a, 0, 1) . lcs(substr(\$a, 1), substr(\$b, 1));
}
my \$c = lcs(substr(\$a, 1), \$b) || "";
my \$d = lcs(\$a, substr(\$b, 1)) || "";
return length(\$c) > length(\$d) ? \$c : \$d;
}

print lcs("thisisatest", "testing123testing") . "\n";
```

## Perl 6

### Recursion

{{works with|rakudo|2015-09-16}} This solution is similar to the Haskell one. It is slow.

```say lcs("thisisatest", "testing123testing");sub lcs(Str \$xstr, Str \$ystr) {
return "" unless \$xstr && \$ystr;

my (\$x, \$xs, \$y, \$ys) = \$xstr.substr(0, 1), \$xstr.substr(1), \$ystr.substr(0, 1), \$ystr.substr(1);
return \$x eq \$y
?? \$x ~ lcs(\$xs, \$ys)
!! max(:by{ \$^a.chars }, lcs(\$xstr, \$ys), lcs(\$xs, \$ystr) );
}

say lcs("thisisatest", "testing123testing");
```

### Dynamic Programming

{{trans|Java}}

```
sub lcs(Str \$xstr, Str \$ystr) {
my (\$xlen, \$ylen) = (\$xstr, \$ystr)>>.chars;
my @lengths = map {[(0) xx (\$ylen+1)]}, 0..\$xlen;

for \$xstr.comb.kv -> \$i, \$x {
for \$ystr.comb.kv -> \$j, \$y {
@lengths[\$i+1][\$j+1] = \$x eq \$y ?? @lengths[\$i][\$j]+1 !! (@lengths[\$i+1][\$j], @lengths[\$i][\$j+1]).max;
}
}

my @x = \$xstr.comb;
my (\$x, \$y) = (\$xlen, \$ylen);
my \$result = "";
while \$x != 0 && \$y != 0 {
if @lengths[\$x][\$y] == @lengths[\$x-1][\$y] {
\$x--;
}
elsif @lengths[\$x][\$y] == @lengths[\$x][\$y-1] {
\$y--;
}
else {
\$result = @x[\$x-1] ~ \$result;
\$x--;
\$y--;
}
}

return \$result;
}

say lcs("thisisatest", "testing123testing");
```

### Bit Vector

Bit parallel dynamic programming with nearly linear complexity O(n). It is fast.

```sub lcs(Str \$xstr, Str \$ystr) {
my (\$a,\$b) = ([\$xstr.comb],[\$ystr.comb]);

my \$positions;
for \$a.kv -> \$i,\$x { \$positions{\$x} +|= 1 +< \$i };

my \$S = +^0;
my \$Vs = [];
my (\$y,\$u);
for (0..+\$b-1) -> \$j {
\$y = \$positions{\$b[\$j]} // 0;
\$u = \$S +& \$y;
\$S = (\$S + \$u) +| (\$S - \$u);
\$Vs[\$j] = \$S;
}

my (\$i,\$j) = (+\$a-1, +\$b-1);
my \$result = "";
while (\$i >= 0 && \$j >= 0) {
if (\$Vs[\$j] +& (1 +< \$i)) { \$i-- }
else {
unless (\$j && +^\$Vs[\$j-1] +& (1 +< \$i)) {
\$result = \$a[\$i] ~ \$result;
\$i--;
}
\$j--;
}
}
return \$result;
}

say lcs("thisisatest", "testing123testing");
```

## Phix

If you want this to work with (utf8) Unicode text, just chuck the inputs through utf8_to_utf32() first (and the output through utf32_to_utf8()).

```function lcs(sequence a, b)
sequence res = ""
if length(a) and length(b) then
if a[\$]=b[\$] then
res = lcs(a[1..-2],b[1..-2])&a[\$]
else
sequence l = lcs(a,b[1..-2]),
r = lcs(a[1..-2],b)
res = iff(length(l)>length(r)?l:r)
end if
end if
return res
end function

constant tests = {{"1234","1224533324"},
{"thisisatest","testing123testing"}}
for i=1 to length(tests) do
string {a,b} = tests[i]
?lcs(a,b)
end for
```

{{out}}

```
"1234"
"tsitest"

```

### Alternate version

same output

```function LCSLength(sequence X, sequence Y)
sequence C = repeat(repeat(0,length(Y)+1),length(X)+1)
for i=1 to length(X) do
for j=1 to length(Y) do
if X[i]=Y[j] then
C[i+1][j+1] := C[i][j]+1
else
C[i+1][j+1] := max(C[i+1][j], C[i][j+1])
end if
end for
end for
return C
end function

function backtrack(sequence C, sequence X, sequence Y, integer i, integer j)
if i=0 or j=0 then
return ""
elsif X[i]=Y[j] then
return backtrack(C, X, Y, i-1, j-1) & X[i]
else
if C[i+1][j]>C[i][j+1] then
return backtrack(C, X, Y, i, j-1)
else
return backtrack(C, X, Y, i-1, j)
end if
end if
end function

function lcs(sequence a, sequence b)
return backtrack(LCSLength(a,b),a,b,length(a),length(b))
end function

constant tests = {{"1234","1224533324"},
{"thisisatest","testing123testing"}}
for i=1 to length(tests) do
string {a,b} = tests[i]
?lcs(a,b)
end for
```

## PicoLisp

```(de commonSequences (A B)
(when A
(conc
(when (member (car A) B)
(mapcar '((L) (cons (car A) L))
(cons NIL (commonSequences (cdr A) (cdr @))) ) )
(commonSequences (cdr A) B) ) ) )

(maxi length
(commonSequences
(chop "thisisatest")
(chop "testing123testing") ) )
```

{{out}}

```-> ("t" "s" "i" "t" "e" "s" "t")
```

## PowerShell

Returns a sequence (array) of a type:

```
function Get-Lcs (\$ReferenceObject, \$DifferenceObject)
{
\$longestCommonSubsequence = @()
\$x = \$ReferenceObject.Length
\$y = \$DifferenceObject.Length

\$lengths = New-Object -TypeName 'System.Object[,]' -ArgumentList (\$x + 1), (\$y + 1)

for(\$i = 0; \$i -lt \$x; \$i++)
{
for (\$j = 0; \$j -lt \$y; \$j++)
{
if (\$ReferenceObject[\$i] -ceq \$DifferenceObject[\$j])
{
\$lengths[(\$i+1),(\$j+1)] = \$lengths[\$i,\$j] + 1
}
else
{
\$lengths[(\$i+1),(\$j+1)] = [Math]::Max((\$lengths[(\$i+1),\$j]),(\$lengths[\$i,(\$j+1)]))
}
}
}

while ((\$x -ne 0) -and (\$y -ne 0))
{
if ( \$lengths[\$x,\$y] -eq \$lengths[(\$x-1),\$y])
{
--\$x
}
elseif (\$lengths[\$x,\$y] -eq \$lengths[\$x,(\$y-1)])
{
--\$y
}
else
{
if (\$ReferenceObject[(\$x-1)] -ceq \$DifferenceObject[(\$y-1)])
{
\$longestCommonSubsequence = ,(\$ReferenceObject[(\$x-1)]) + \$longestCommonSubsequence
}

--\$x
--\$y
}
}

\$longestCommonSubsequence
}

```

Returns the character array as a string:

```
(Get-Lcs -ReferenceObject "thisisatest" -DifferenceObject "testing123testing") -join ""

```

{{Out}}

```
tsitest

```

Returns an array of integers:

```
Get-Lcs -ReferenceObject @(1,2,3,4) -DifferenceObject @(1,2,2,4,5,3,3,3,2,4)

```

{{Out}}

```
1
2
3
4

```

Given two lists of objects, return the LCS of the ID property:

```
\$list1

ID   X   Y
--   -   -
1 101 201
2 102 202
3 103 203
4 104 204
5 105 205
6 106 206
7 107 207
8 108 208
9 109 209

\$list2

ID   X   Y
--   -   -
1 101 201
3 103 203
5 105 205
7 107 207
9 109 209

Get-Lcs -ReferenceObject \$list1.ID -DifferenceObject \$list2.ID

```

{{Out}}

```
1
3
5
7
9

```

## Prolog

### Recursive Version

First version:

```test :-
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).

lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).

lcs([H1|L1],[H2|L2],Lcs):-
lcs(    L1 ,[H2|L2],Lcs1),
lcs([H1|L1],    L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!.

lcs(_,_,[]).

longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).
```

Second version, with memoization:

```%declare that we will add lcs_db facts during runtime
:- dynamic lcs_db/3.

test :-
retractall(lcs_db(_,_,_)), %clear the database of known results
time(lcs("thisisatest", "testing123testing", Lcs)),
writef('%s',[Lcs]).

% check if the result is known
lcs(L1,L2,Lcs) :-
lcs_db(L1,L2,Lcs),!.

lcs([ H|L1],[ H|L2],[H|Lcs]) :- !,
lcs(L1,L2,Lcs).

lcs([H1|L1],[H2|L2],Lcs) :-
lcs(    L1 ,[H2|L2],Lcs1),
lcs([H1|L1],    L2 ,Lcs2),
longest(Lcs1,Lcs2,Lcs),!,
assert(lcs_db([H1|L1],[H2|L2],Lcs)).

lcs(_,_,[]).

longest(L1,L2,Longest) :-
length(L1,Length1),
length(L2,Length2),
((Length1 > Length2) -> Longest = L1; Longest = L2).
```

{{out|Demonstrating}} Example for "beginning-middle-ending" and "beginning-diddle-dum-ending"
First version :

```?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 10,875,184 inferences, 1.840 CPU in 1.996 seconds (92% CPU, 5910426 Lips)
beginning-iddle-ending
```

Second version which is much faster :

```?- time(lcs("beginning-middle-ending","beginning-diddle-dum-ending", Lcs)),writef('%s', [Lcs]).
% 2,376 inferences, 0.010 CPU in 0.003 seconds (300% CPU, 237600 Lips)
beginning-iddle-ending
```

## PureBasic

{{trans|Basic}}

```Procedure.s lcs(a\$, b\$)
Protected x\$ , lcs\$
If Len(a\$) = 0 Or Len(b\$) = 0
lcs\$ = ""
ElseIf Right(a\$, 1) = Right(b\$, 1)
lcs\$ = lcs(Left(a\$, Len(a\$) - 1), Left(b\$, Len(b\$) - 1)) + Right(a\$, 1)
Else
x\$ = lcs(a\$, Left(b\$, Len(b\$) - 1))
y\$ = lcs(Left(a\$, Len(a\$) - 1), b\$)
If Len(x\$) > Len(y\$)
lcs\$ = x\$
Else
lcs\$ = y\$
EndIf
EndIf
ProcedureReturn lcs\$
EndProcedure
OpenConsole()
PrintN( lcs("thisisatest", "testing123testing"))
PrintN("Press any key to exit"): Repeat: Until Inkey() <> ""
```

## Python

The simplest way is to use [http://mlpy.sourceforge.net/docs/3.5/lcs.html LCS within mlpy package]

### Recursion

This solution is similar to the Haskell one. It is slow.

```def lcs(xstr, ystr):
"""
>>> lcs('thisisatest', 'testing123testing')
'tsitest'
"""
if not xstr or not ystr:
return ""
x, xs, y, ys = xstr[0], xstr[1:], ystr[0], ystr[1:]
if x == y:
return x + lcs(xs, ys)
else:
return max(lcs(xstr, ys), lcs(xs, ystr), key=len)
```

Test it:

```if __name__=="__main__":
import doctest; doctest.testmod()
```

### Dynamic Programming

```def lcs(a, b):
# generate matrix of length of longest common subsequence for substrings of both words
lengths = [[0] * (len(b)+1) for _ in range(len(a)+1)]
for i, x in enumerate(a):
for j, y in enumerate(b):
if x == y:
lengths[i+1][j+1] = lengths[i][j] + 1
else:
lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1])

# read a substring from the matrix
result = ''
j = len(b)
for i in range(1, len(a)+1):
if lengths[i][j] != lengths[i-1][j]:
result += a[i-1]

return result
```

## Racket

```#lang racket
(define (longest xs ys)
(if (> (length xs) (length ys))
xs ys))

(define memo (make-hash))
(define (lookup xs ys)
(hash-ref memo (cons xs ys) #f))
(define (store xs ys r)
(hash-set! memo (cons xs ys) r)
r)

(define (lcs/list sx sy)
(or (lookup sx sy)
(store sx sy
(match* (sx sy)
[((cons x xs) (cons y ys))
(if (equal? x y)
(cons x (lcs/list xs ys))
(longest (lcs/list sx ys) (lcs/list xs sy)))]
[(_ _) '()]))))

(define (lcs sx sy)
(list->string (lcs/list (string->list sx) (string->list sy))))

(lcs "thisisatest" "testing123testing")
```

{{out}}

```"tsitest">
```

## REXX

```/*REXX program to test the  LCS (Longest Common Subsequence) subroutine.*/
parse arg aaa bbb .                    /*get two arguments (strings).   */
say 'string A = 'aaa                   /*echo string  A  to screen.     */
say 'string B = 'bbb                   /*echo string  B  to screen.     */
say '     LCS = 'lcs(aaa,bbb)          /*tell Longest Common Sequence.  */
exit                                   /*stick a fork in it, we're done.*/
/*──────────────────────────────────LCS subroutine──────────────────────*/
lcs: procedure; parse arg a,b,z        /*Longest Common Subsequence.    */
/*reduce recursions, removes the */
/*chars in A ¬ in B, & vice-versa*/
if z=='' then return lcs( lcs(a,b,0), lcs(b,a,0), 9)
j=length(a)
if z==0 then do                        /*special invocation:  shrink Z. */
do j=1  for j;   _=substr(a,j,1)
if pos(_,b)\==0  then z=z||_
end   /*j*/
return substr(z,2)
end
k=length(b)
if j==0 | k==0  then return ''         /*Either string null?    Bupkis. */
_=substr(a,j,1)
if _==substr(b,k,1)  then return lcs(substr(a,1,j-1),substr(b,1,k-1),9)_
x=lcs(a,substr(b,1,k-1),9)
y=lcs(substr(a,1,j-1),b,9)
if length(x)>length(y)  then return x
return y
```

{{out|Output with input “ 1234 1224533324 ”}}

```
string A=1234
string B=1224533324
LCS=1234

```

{{out|Output with input “ thisisatest testing123testing ”}}

```
string A=thisisatest
string B=testing123testing
LCS=tsitest

```

## Ring

```
see longest("1267834", "1224533324") + nl

func longest a, b
if a = "" or b = "" return "" ok
if right(a, 1) = right(b, 1)
lcs = longest(left(a, len(a) - 1), left(b, len(b) - 1)) + right(a, 1)
return lcs
else
x1 = longest(a, left(b, len(b) - 1))
x2 = longest(left(a, len(a) - 1), b)
if len(x1) > len(x2)
lcs = x1
return lcs
else
lcs = x2
return lcs ok ok

```

Output:

```
1234

```

## Ruby

### Recursion

This solution is similar to the Haskell one. It is slow (The time complexity is exponential.) {{works with|Ruby|1.9}}

```=begin
irb(main):001:0> lcs('thisisatest', 'testing123testing')
=> "tsitest"
=end
def lcs(xstr, ystr)
return "" if xstr.empty? || ystr.empty?

x, xs, y, ys = xstr[0..0], xstr[1..-1], ystr[0..0], ystr[1..-1]
if x == y
x + lcs(xs, ys)
else
[lcs(xstr, ys), lcs(xs, ystr)].max_by {|x| x.size}
end
end
```

### Dynamic programming

{{works with|Ruby|1.9}}

Walker class for the LCS matrix:

```class LCS
SELF, LEFT, UP, DIAG = [0,0], [0,-1], [-1,0], [-1,-1]

def initialize(a, b)
@m = Array.new(a.length) { Array.new(b.length) }
a.each_char.with_index do |x, i|
b.each_char.with_index do |y, j|
match(x, y, i, j)
end
end
end

def match(c, d, i, j)
@i, @j = i, j
@m[i][j] = compute_entry(c, d)
end

def lookup(x, y)        [@i+x, @j+y]                      end
def valid?(i=@i, j=@j)  i >= 0 && j >= 0                  end

def peek(x, y)
i, j = lookup(x, y)
valid?(i, j) ? @m[i][j] : 0
end

def compute_entry(c, d)
c == d ? peek(*DIAG) + 1 : [peek(*LEFT), peek(*UP)].max
end

def backtrack
@i, @j = @m.length-1, @m[0].length-1
y = []
y << @i+1 if backstep? while valid?
y.reverse
end

def backtrack2
@i, @j = @m.length-1, @m[0].length-1
y = []
y << @j+1 if backstep? while valid?
[backtrack, y.reverse]
end

def backstep?
backstep = compute_backstep
@i, @j = lookup(*backstep)
backstep == DIAG
end

def compute_backstep
case peek(*SELF)
when peek(*LEFT) then LEFT
when peek(*UP)   then UP
else                  DIAG
end
end
end

def lcs(a, b)
walker = LCS.new(a, b)
walker.backtrack.map{|i| a[i]}.join
end

if \$0 == __FILE__
puts lcs('thisisatest', 'testing123testing')
puts lcs("rosettacode", "raisethysword")
end
```

{{out}}

```
tsitest
rsetod

```

Referring to LCS [[Levenshtein distance/Alignment#Ruby|here]] and [[Shortest common supersequence#Ruby|here]].

## Run BASIC

```a\$	= "aebdaef"
b\$	= "cacbac"
print lcs\$(a\$,b\$)
end

FUNCTION lcs\$(a\$, b\$)
IF a\$ = "" OR b\$ = "" THEN
lcs\$ = ""
goto [ext]
end if

IF RIGHT\$(a\$, 1) = RIGHT\$(b\$, 1) THEN
lcs\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), LEFT\$(b\$, LEN(b\$) - 1)) + RIGHT\$(a\$, 1)
goto [ext]
ELSE
x1\$ = lcs\$(a\$, LEFT\$(b\$, LEN(b\$) - 1))
x2\$ = lcs\$(LEFT\$(a\$, LEN(a\$) - 1), b\$)
IF LEN(x1\$) > LEN(x2\$) THEN
lcs\$ = x1\$
goto [ext]
ELSE
lcs\$ = x2\$
goto [ext]
END IF
END IF
[ext]
END FUNCTION
```
```aba
```

## Rust

Dynamic programming version:

```
use std::cmp;

fn lcs(string1: String, string2: String) -> (usize, String){
let total_rows = string1.len() + 1;
let total_columns = string2.len() + 1;
// rust doesn't allow accessing string by index
let string1_chars = string1.as_bytes();
let string2_chars = string2.as_bytes();

let mut table = vec![vec![0; total_columns]; total_rows];

for row in 1..total_rows{
for col in 1..total_columns {
if string1_chars[row - 1] == string2_chars[col - 1]{
table[row][col] = table[row - 1][col - 1] + 1;
} else {
table[row][col] = cmp::max(table[row][col-1], table[row-1][col]);
}
}
}

let mut common_seq = Vec::new();
let mut x = total_rows - 1;
let mut y = total_columns - 1;

while x != 0 && y != 0 {
// Check element above is equal
if table[x][y] == table[x - 1][y] {
x = x - 1;
}
// check element to the left is equal
else if table[x][y] == table[x][y - 1] {
y = y - 1;
}
else {
// check the two element at the respective x,y position is same
assert_eq!(string1_chars[x-1], string2_chars[y-1]);
let char = string1_chars[x - 1];
common_seq.push(char);
x = x - 1;
y = y - 1;
}
}
common_seq.reverse();
(table[total_rows - 1][total_columns - 1], String::from_utf8(common_seq).unwrap())
}

fn main() {
let res = lcs("abcdaf".to_string(), "acbcf".to_string());
assert_eq!((4 as usize, "abcf".to_string()), res);
let res = lcs("thisisatest".to_string(), "testing123testing".to_string());
assert_eq!((7 as usize, "tsitest".to_string()), res);
// LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4.
let res = lcs("AGGTAB".to_string(), "GXTXAYB".to_string());
assert_eq!((4 as usize, "GTAB".to_string()), res);
}
```

## Scala

{{works with|Scala 2.13}} Using lazily evaluated lists:

```  def lcsLazy[T](u: IndexedSeq[T], v: IndexedSeq[T]): IndexedSeq[T] = {
def su = subsets(u).to(LazyList)
def sv = subsets(v).to(LazyList)
case Some(sub) => sub
case None => IndexedSeq[T]()
}
}

def subsets[T](u: IndexedSeq[T]): Iterator[IndexedSeq[T]] = {
u.indices.reverseIterator.flatMap{n => u.indices.combinations(n + 1).map(_.map(u))}
}
```

Using recursion:

```  def lcsRec[T]: (IndexedSeq[T], IndexedSeq[T]) => IndexedSeq[T] = {
case (a +: as, b +: bs) if a == b => a +: lcsRec(as, bs)
case (as, bs) if as.isEmpty || bs.isEmpty => IndexedSeq[T]()
case (a +: as, b +: bs) =>
val (s1, s2) = (lcsRec(a +: as, bs), lcsRec(as, b +: bs))
if(s1.length > s2.length) s1 else s2
}
```

{{out}}

```scala> lcsLazy("thisisatest", "testing123testing").mkString
res0: String = tsitest

scala> lcsRec("thisisatest", "testing123testing").mkString
res1: String = tsitest
```

{{works with|Scala 2.9.3}} Recursive version:

```  def lcs[T]: (List[T], List[T]) => List[T] = {
case (_, Nil) => Nil
case (Nil, _) => Nil
case (x :: xs, y :: ys) if x == y => x :: lcs(xs, ys)
case (x :: xs, y :: ys)           => {
(lcs(x :: xs, ys), lcs(xs, y :: ys)) match {
case (xs, ys) if xs.length > ys.length => xs
case (xs, ys)                          => ys
}
}
}
```

The dynamic programming version:

```  case class Memoized[A1, A2, B](f: (A1, A2) => B) extends ((A1, A2) => B) {
val cache = scala.collection.mutable.Map.empty[(A1, A2), B]
def apply(x: A1, y: A2) = cache.getOrElseUpdate((x, y), f(x, y))
}

lazy val lcsM: Memoized[List[Char], List[Char], List[Char]] = Memoized {
case (_, Nil) => Nil
case (Nil, _) => Nil
case (x :: xs, y :: ys) if x == y => x :: lcsM(xs, ys)
case (x :: xs, y :: ys)           => {
(lcsM(x :: xs, ys), lcsM(xs, y :: ys)) match {
case (xs, ys) if xs.length > ys.length => xs
case (xs, ys)                          => ys
}
}
}
```

{{out}} scala> lcsM("thisiaatest".toList, "testing123testing".toList).mkString res0: String = tsitest

## Scheme

Port from Clojure.

```
;; using srfi-69
(define (memoize proc)
(let ((results (make-hash-table)))
(lambda args
(or (hash-table-ref results args (lambda () #f))
(let ((r (apply proc args)))
(hash-table-set! results args r)
r)))))

(define (longest xs ys)
(if (> (length xs)
(length ys))
xs ys))

(define lcs
(memoize
(lambda (seqx seqy)
(if (pair? seqx)
(let ((x (car seqx))
(xs (cdr seqx)))
(if (pair? seqy)
(let ((y (car seqy))
(ys (cdr seqy)))
(if (equal? x y)
(cons x (lcs xs ys))
(longest (lcs seqx ys)
(lcs xs seqy))))
'()))
'()))))

```

Testing:

```

(test-group
"lcs"
(test '()  (lcs '(a b c) '(A B C)))
(test '(a) (lcs '(a a a) '(A A a)))
(test '()  (lcs '() '(a b c)))
(test '()  (lcs '(a b c) '()))
(test '(a c) (lcs '(a b c) '(a B c)))
(test '(b) (lcs '(a b c) '(A b C)))

(test     '(  b   d e f     g h   j)
(lcs '(a b   d e f     g h i j)
'(A b c d e f F a g h   j))))

```

## Seed7

```\$ include "seed7_05.s7i";

const func string: lcs (in string: a, in string: b) is func
result
var string: lcs is "";
local
var string: x is "";
var string: y is "";
begin
if a <> "" and b <> "" then
if a[length(a)] = b[length(b)] then
lcs := lcs(a[.. pred(length(a))], b[.. pred(length(b))]) & str(a[length(a)]);
else
x := lcs(a, b[.. pred(length(b))]);
y := lcs(a[.. pred(length(a))], b);
if length(x) > length(y) then
lcs := x;
else
lcs := y;
end if;
end if;
end if;
end func;

const proc: main is func
begin
writeln(lcs("thisisatest", "testing123testing"));
writeln(lcs("1234", "1224533324"));
end func;
```

Output:

```
tsitest
1234

```

## SequenceL

{{trans|C#}}

It is interesting to note that x and y are computed in parallel, dividing work across threads repeatedly down through the recursion.

```;

lcsBack(a(1), b(1)) :=
let
aSub := allButLast(a);
bSub := allButLast(b);

x := lcsBack(a, bSub);
y := lcsBack(aSub, b);
in
[] when size(a) = 0 or size(b) = 0
else
lcsBack(aSub, bSub) ++ [last(a)] when last(a) = last(b)
else
x when size(x) > size(y)
else
y;

main(args(2)) :=
lcsBack(args[1], args[2]) when size(args) >=2
else
lcsBack("thisisatest", "testing123testing");
```

{{out}}

```
"tsitest"

```

## SETL

Recursive; Also works on tuples (vectors)

```    op .longest(a, b);
return if #a > #b then a else b end;
end .longest;

procedure lcs(a, b);
if exists empty in {a, b} | #empty = 0 then
return empty;
elseif a(1) = b(1) then
return a(1) + lcs(a(2..), b(2..));
else
return lcs(a(2..), b) .longest lcs(a, b(2..));
end;
end lcs;
```

## Sidef

```func lcs(xstr, ystr) is cached {

xstr.is_empty && return xstr;
ystr.is_empty && return ystr;

var(x, xs, y, ys) = (xstr.ft(0,0), xstr.ft(1),
ystr.ft(0,0), ystr.ft(1));

if (x == y) {
x + lcs(xs, ys)
} else {
[lcs(xstr, ys), lcs(xs, ystr)].max_by { .len };
}
}

say lcs("thisisatest", "testing123testing");
```

{{out}}

```
tsitest

```

## Slate

We define this on the Sequence type since there is nothing string-specific about the concept.

### Recursion

{{trans|Java}}

```s1@(Sequence traits) longestCommonSubsequenceWith: s2@(Sequence traits)
[
s1 isEmpty \/ s2 isEmpty ifTrue: [^ {}].
s1 last = s2 last
ifTrue: [(s1 allButLast longestCommonSubsequenceWith: s2 allButLast) copyWith: s1 last]
ifFalse: [| x y |
x: (s1 longestCommonSubsequenceWith: s2 allButLast).
y: (s1 allButLast longestCommonSubsequenceWith: s2).
x length > y length ifTrue: [x] ifFalse: [y]]
].
```

### Dynamic Programming

{{trans|Ruby}}

```s1@(Sequence traits) longestCommonSubsequenceWith: s2@(Sequence traits)
[| lengths |
lengths: (ArrayMD newWithDimensions: {s1 length `cache. s2 length `cache} defaultElement: 0).
s1 doWithIndex: [| :elem1 :index1 |
s2 doWithIndex: [| :elem2 :index2 |
elem1 = elem2
ifTrue: [lengths at: {index1 + 1. index2 + 1} put: (lengths at: {index1. index2}) + 1]
ifFalse: [lengths at: {index1 + 1. index2 + 1} put:
((lengths at: {index1 + 1. index2}) max: (lengths at: {index1. index2 + 1}))]]].
([| :result index1 index2 |
index1: s1 length.
index2: s2 length.
[index1 isPositive /\ index2 isPositive]
whileTrue:
[(lengths at: {index1. index2}) = (lengths at: {index1 - 1. index2})
ifTrue: [index1: index1 - 1]
ifFalse: [(lengths at: {index1. index2}) = (lengths at: {index1. index2 - 1})]
ifTrue: [index2: index2 - 1]
ifFalse: ["assert: (s1 at: index1 - 1) = (s2 at: index2 - 1)."
result nextPut: (s1 at: index1 - 1).
index1: index1 - 1.
index2: index2 - 1]]
] writingAs: s1) reverse
].
```

## Swift

{{trans|Java}} Swift 2.2

### Recursion

```func rlcs(_ s1: String, _ s2: String) -> String {
let x = s1.characters.count
let y = s2.characters.count

if x == 0 || y == 0 {
return ""
} else if s1[s1.index(s1.startIndex, offsetBy: x-1)] == s2[s2.index(s2.startIndex, offsetBy: y-1)] {
return rlcs(String(s1[s1.startIndex..<s1.index(s1.startIndex, offsetBy: x-1)]),
String(s2[s2.startIndex..<s2.index(s2.startIndex, offsetBy: y-1)])) + String(s1[s1.index(s1.startIndex, offsetBy: x-1)])
} else {
let xstr = rlcs(s1, String(s2[s2.startIndex..<s2.index(s2.startIndex, offsetBy: y-1)]))
let ystr = rlcs(String(s1[s1.startIndex..<s1.index(s1.startIndex, offsetBy: x-1)]), s2)

return xstr.characters.count > ystr.characters.count ? xstr : ystr
}
}
```

### Dynamic Programming

```func lcs(s1:String, _ s2:String) -> String {
var x = s1.characters.count
var y = s2.characters.count
var lens = Array(count: x + 1, repeatedValue:
Array(count: y + 1, repeatedValue: 0))
var returnStr = ""

for i in 0..<x {
for j in 0..<y {
lens[i + 1][j + 1] = lens[i][j] + 1
} else {
lens[i + 1][j + 1] = max(lens[i + 1][j], lens[i][j + 1])
}
}
}

while x != 0 && y != 0 {
if lens[x][y] == lens[x - 1][y] {
--x
} else if lens[x][y] == lens[x][y - 1] {
--y
} else {
--x
--y
}
}

return String(returnStr.characters.reverse())
}
```

## Tcl

### Recursive

{{trans|Java}}

```proc r_lcs {a b} {
if {\$a eq "" || \$b eq ""} {return ""}
set a_ [string range \$a 1 end]
set b_ [string range \$b 1 end]
if {[set c [string index \$a 0]] eq [string index \$b 0]} {
return "\$c[r_lcs \$a_ \$b_]"
} else {
set x [r_lcs \$a \$b_]
set y [r_lcs \$a_ \$b]
return [expr {[string length \$x] > [string length \$y] ? \$x :\$y}]
}
}
```

### Dynamic

{{trans|Java}} {{works with|Tcl|8.5}}

```package require Tcl 8.5
namespace import ::tcl::mathop::+
namespace import ::tcl::mathop::-
namespace import ::tcl::mathfunc::max

proc d_lcs {a b} {
set la [string length \$a]
set lb [string length \$b]
set lengths [lrepeat [+ \$la 1] [lrepeat [+ \$lb 1] 0]]

for {set i 0} {\$i < \$la} {incr i} {
for {set j 0} {\$j < \$lb} {incr j} {
if {[string index \$a \$i] eq [string index \$b \$j]} {
lset lengths [+ \$i 1] [+ \$j 1] [+ [lindex \$lengths \$i \$j] 1]
} else {
lset lengths [+ \$i 1] [+ \$j 1] [max [lindex \$lengths [+ \$i 1] \$j] [lindex \$lengths \$i [+ \$j 1]]]
}
}
}

set result ""
set x \$la
set y \$lb
while {\$x >0 && \$x > 0} {
if {[lindex \$lengths \$x \$y] == [lindex \$lengths [- \$x 1] \$y]} {
incr x -1
} elseif {[lindex \$lengths \$x \$y] == [lindex \$lengths \$x [- \$y 1]]} {
incr y -1
} else {
if {[set c [string index \$a [- \$x 1]]] ne [string index \$b [- \$y 1]]} {
error "assertion failed: a.charAt(x-1) == b.charAt(y-1)"
}
append result \$c
incr x -1
incr y -1
}
}
return [string reverse \$result]
}
```

### Performance Comparison

```% time {d_lcs thisisatest testing123testing} 10
637.5 microseconds per iteration
% time {r_lcs thisisatest testing123testing} 10
1275566.8 microseconds per iteration
```

## Ursala

This uses the same recursive algorithm as in the Haskell example, and works on lists of any type.

```#import std

lcs = ~&alrB^& ~&E?abh/~&alh2fabt2RC @faltPrXlrtPXXPW leql?/~&r ~&l
```

test program:

```#cast %s

example = lcs('thisisatest','testing123testing')
```

{{out}}

```'tsitest'
```

## zkl

This is quite vile in terms of [time] efficiency, another algorithm should be used for real work. {{trans|D}}

```fcn lcs(a,b){
if(not a or not b) return("");
if (a[0]==b[0]) return(a[0] + self.fcn(a[1,*],b[1,*]));
return(fcn(x,y){if(x.len()>y.len())x else y}(lcs(a,b[1,*]),lcs(a[1,*],b)))
}
```

The last line looks strange but it is just return(lambda longest(lcs.lcs)) {{out}}

```
zkl: lcs("thisisatest", "testing123testing")
tsitest

```