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Introduction

Define a subsequence to be any output string obtained by deleting zero or more symbols from an input string.

The Longest Common Subsequence (LCS) is a subsequence of maximum length common to two or more strings.

Let A ≡ A[0]… A[m - 1] and B ≡ B[0]… B[n - 1], m < n be strings drawn from an alphabet Σ of size s, containing every distinct symbol in A + B.

An ordered pair (i, j) will be referred to as a match if A[i] = B[j], where 0 ≤ i < m and 0 ≤ j < n.

The set of matches M defines a relation over matches: M[i, j] ⇔ (i, j) ∈ M.

Define a non-strict product-order (≤) over ordered pairs, such that (i1, j1) ≤ (i2, j2) ⇔ i1 ≤ i2 and j1 ≤ j2. We define (≥) similarly.

We say ordered pairs p1 and p2 are comparable if either p1 ≤ p2 or p1 ≥ p2 holds. If i1 < i2 and j2 < j1 (or i2 < i1 and j1 < j2) then neither p1 ≤ p2 nor p1 ≥ p2 are possible, and we say p1 and p2 are incomparable.

Define the strict product-order (<) over ordered pairs, such that (i1, j1) < (i2, j2) ⇔ i1 < i2 and j1 < j2. We define (>) similarly.

A chain C is a subset of M consisting of at least one element m; and where either m1 < m2 or m1 > m2 for every pair of distinct elements m1 and m2. An antichain D is any subset of M in which every pair of distinct elements m1 and m2 are incomparable.

A chain can be visualized as a strictly increasing curve that passes through matches (i, j) in the m*n coordinate space of M[i, j].

Every Common Sequence of length q corresponds to a chain of cardinality q, over the set of matches M. Thus, finding an LCS can be restated as the problem of finding a chain of maximum cardinality p.

According to [Dilworth 1950], this cardinality p equals the minimum number of disjoint antichains into which M can be decomposed. Note that such a decomposition into the minimal number p of disjoint antichains may not be unique.

Background

Where the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols increases; and the number of matches will tend towards O(m*n) quadratic growth. This occurs, for example, in the Bioinformatics application of nucleotide and protein sequencing.

The divide-and-conquer approach of [Hirschberg 1975] limits the space required to O(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 approaches the length of the LCS. In this case the number of matches reduces to linear, O(n) growth.

A binary search optimization due to [Hunt and Szymanski 1977] can be applied to the basic Dynamic Programming approach, resulting in an expected performance of O(n log m). Performance can degrade to O(m*n log m) time in the worst case, as the number of matches grows to O(m*n).

Note

[Rick 2000] describes a linear-space algorithm with a time bound of O(n*s + p*min(m, n - p)).

Legend

A, B are input strings of lengths m, n respectively
p is the length of the LCS
M is the set of matches (i, j) such that A[i] = B[j]
r is the magnitude of M
s is the magnitude of the alphabet Σ of distinct symbols in A + B

References

[Dilworth 1950] "A decomposition theorem for partially ordered sets" by Robert P. Dilworth, published January 1950, Annals of Mathematics [Volume 51, Number 1, pp. 161-166]

[Goeman and Clausen 2002] "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]

[Hirschberg 1975] "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]

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

[Hunt and Szymanski 1977] "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]

[Rick 2000] "Simple and fast linear space computation of longest common subsequences" by Claus Rick, received 17 March 2000, Information Processing Letters, Elsevier Science [Volume 75, pp. 275–281]

Examples

The sequences "1234" and "1224533324" have an LCS of "1234":

1234
1224533324

For a string example, consider the sequences "thisisatest" and "testing123testing". An LCS would be "tsitest":

thisisatest
testing123testing

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.

For more information on this problem please see Wikipedia.

Other tasks related to string operations:

Metrics

Counting

Remove/replace

Anagrams/Derangements/shuffling

Find/Search/Determine

Formatting

Song lyrics/poems/Mad Libs/phrases

Tokenize

Sequences

11l

Translation of: Python

F lcs(a, b)
   V lengths = [[0] * (b.len+1)] * (a.len+1)
   L(x) a
      V i = L.index
      L(y) b
         V j = L.index
         I x == y
            lengths[i + 1][j + 1] = lengths[i][j] + 1
         E
            lengths[i + 1][j + 1] = max(lengths[i + 1][j], lengths[i][j + 1])

   V result = ‘’
   V j = b.len
   L(i) 1..a.len
      I lengths[i][j] != lengths[i - 1][j]
         result ‘’= a[i - 1]
   R result

print(lcs(‘1234’, ‘1224533324’))
print(lcs(‘thisisatest’, ‘testing123testing’))

Output:

1234
tisitst

Ada

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;

Output:

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;

Output:

tsitest

ALGOL 68

Translation of: Ada

Works with: ALGOL 68 version Standard - no extensions to language used

Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386

Works with: ELLA ALGOL 68 version 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))
)

Output:

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
 }

Arturo

Translation of: Python

lcs: function [a,b][
    ls: new array.of: @[inc size a, inc size b] 0

    loop.with:'i a 'x [
        loop.with:'j b 'y [
            ls\[i+1]\[j+1]: (x=y)? -> ls\[i]\[j] + 1
                                   -> max @[ls\[i+1]\[j], ls\[i]\[j+1]]
        ]
    ]
    [result, x, y]: @[new "", size a, size b]

    while [and? [x > 0][y > 0]][
        if? ls\[x]\[y] = ls\[x-1]\[y] -> x: x-1
        else [
            if? ls\[x]\[y] = ls\[x]\[y-1] -> y: y-1
            else [
                result: a\[x-1] ++ result
                x: x-1
                y: y-1
            ]
        ]
    ]
    return result
]
 
print lcs "1234" "1224533324"
print lcs "thisisatest" "testing123testing"

Output:

1234
tsitest

AutoHotkey

Translation of: Java

using dynamic programming

ahk forum: 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

QuickBASIC

Works with: QuickBasic version 4.5

Translation of: 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

BASIC256

Translation of: FreeBASIC

function LCS(a, b)
	if length(a) = 0 or length(b) = 0 then return ""
	if right(a, 1) = right(b, 1) then
		LCS = LCS(left(a, length(a) - 1), left(b, length(b) - 1)) + right(a, 1)
	else
		x = LCS(a, left(b, length(b) - 1))
		y = LCS(left(a, length(a) - 1), b)
		if length(x) > length(y) then return x else return y
	end if
end function

print LCS("1234", "1224533324")
print LCS("thisisatest", "testing123testing")
end

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

BQN

It's easier and faster to get only the length of the longest common subsequence, using LcsLen ← ¯1 ⊑ 0¨∘⊢ {𝕩⌈⌈`𝕨+»𝕩}˝ =⌜⟜⌽. This function can be expanded by changing ⌈ to ⊣⍟(>○≠) (choosing from two arguments one that has the greatest length), and replacing the empty length 0 with the empty string "" in the right places.

LCS ← ¯1 ⊑ "" <⊸∾ ""¨∘⊢ ⊣⍟(>○≠){𝕩𝔽¨𝔽`𝕨∾¨""<⊸»𝕩}˝ (=⌜⥊¨⊣)⟜⌽

Output:

   "1234" LCS "1224533324"
"1234"
   "thisisatest" LCS "testing123testing"
"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);

Output:

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#

With recursion

using System;

namespace LCS
{
    class Program
    {
        static void Main(string[] args)
        {
            string word1 = "thisisatest";
            string word2 = "testing123testing";
            
            Console.WriteLine(lcsBack(word1, word2));
            Console.ReadKey();
        }

        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;
            }
        }
    }
}

C++

Hunt and Szymanski algorithm

#include <stdint.h>
#include <string>
#include <memory>                       // for shared_ptr<>
#include <iostream>
#include <deque>
#include <unordered_map>                //[C++11]
#include <algorithm>                    // for lower_bound()
#include <iterator>                     // for next() and prev()

using namespace std;

class LCS {
protected:
  // Instances of the Pair linked list class are used to recover the LCS:
  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) {
      shared_ptr<Pair> head = nullptr;
      for (auto next = pairs; next != nullptr; next = next->next)
        head = make_shared<Pair>(next->index1, next->index2, head);
      return head;
    }
  };

  typedef deque<shared_ptr<Pair>> PAIRS;
  typedef deque<uint32_t> INDEXES;
  typedef unordered_map<char, INDEXES> CHAR_TO_INDEXES_MAP;
  typedef deque<INDEXES*> MATCHES;

  static uint32_t FindLCS(
    MATCHES& indexesOf2MatchedByIndex1, shared_ptr<Pair>* pairs) {
    auto traceLCS = pairs != nullptr;
    PAIRS chains;
    INDEXES prefixEnd;

    //
    //[Assert]After each index1 iteration prefixEnd[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 : indexesOf2MatchedByIndex1) {
      auto dq2 = *it1;
      auto limit = prefixEnd.end();
      for (auto it2 = dq2.rbegin(); it2 != dq2.rend(); it2++) {
        // Each index1, index2 pair corresponds to a match
        auto index2 = *it2;

        //
        // Note: The reverse iterator it2 visits index2 values in descending order,
        // allowing in-place update of prefixEnd[].  std::lower_bound() is used to
        // perform a binary search.
        //
        limit = lower_bound(prefixEnd.begin(), limit, index2);

        //
        // Look ahead to the next index2 value to optimize Pairs used by the Hunt
        // and Szymanski algorithm.  If the next index2 is also an improvement on
        // the value currently held in prefixEnd[index3], a new Pair will only be
        // superseded on the next index2 iteration.
        //
        // Verify that a next index2 value exists; and that this value is greater
        // than the final index2 value of the LCS prefix at prev(limit):
        //
        auto preferNextIndex2 = next(it2) != dq2.rend() &&
          (limit == prefixEnd.begin() || *prev(limit) < *next(it2));

        //
        // Depending on match redundancy, this optimization may reduce the number
        // of Pair allocations by factors ranging from 2 up to 10 or more.
        //
        if (preferNextIndex2) continue;

        auto index3 = distance(prefixEnd.begin(), limit);

        if (limit == prefixEnd.end()) {
          // Insert Case
          prefixEnd.push_back(index2);
          // Refresh limit iterator:
          limit = prev(prefixEnd.end());
          if (traceLCS) {
            chains.push_back(pushPair(chains, index3, index1, index2));
          }
        }
        else if (index2 < *limit) {
          // Update Case
          // Update limit value:
          *limit = index2;
          if (traceLCS) {
            chains[index3] = pushPair(chains, index3, index1, index2);
          }
        }
      }                                   // next index2

      index1++;
    }                                     // next index1

    if (traceLCS) {
      // Return the LCS as a linked list of matched index pairs:
      auto last = chains.empty() ? nullptr : chains.back();
      // Reverse longest chain
      *pairs = Pair::Reverse(last);
    }

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

private:
  static shared_ptr<Pair> pushPair(
    PAIRS& chains, const ptrdiff_t& index3, uint32_t& index1, uint32_t& index2) {
    auto prefix = index3 > 0 ? chains[index3 - 1] : nullptr;
    return make_shared<Pair>(index1, index2, prefix);
  }

protected:
  //
  // Match() avoids m*n comparisons by using CHAR_TO_INDEXES_MAP 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.
  //
  static void Match(
    CHAR_TO_INDEXES_MAP& indexesOf2MatchedByChar, MATCHES& indexesOf2MatchedByIndex1,
    const string& s1, const string& s2) {
    uint32_t index = 0;
    for (const auto& it : s2)
      indexesOf2MatchedByChar[it].push_back(index++);

    for (const auto& it : s1) {
      auto& dq2 = indexesOf2MatchedByChar[it];
      indexesOf2MatchedByIndex1.push_back(&dq2);
    }
  }

  static 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:
  static string Correspondence(const string& s1, const string& s2) {
    CHAR_TO_INDEXES_MAP indexesOf2MatchedByChar;
    MATCHES indexesOf2MatchedByIndex1;  // holds references into indexesOf2MatchedByChar
    Match(indexesOf2MatchedByChar, indexesOf2MatchedByIndex1, s1, s2);
    shared_ptr<Pair> pairs;             // obtain the LCS as index pairs
    auto length = FindLCS(indexesOf2MatchedByIndex1, &pairs);
    return Select(pairs, length, false, s1, s2);
  }
};

Example:

    auto s = LCS::Correspondence(s1, s2);
    cout << s << endl;

More fully featured examples are available at Samples/C++/LCS.

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))
                             ;; if they start with the same element,
                             ;; 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;
}

Output:

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.

Adapted from Python code: http://wordaligned.org/articles/longest-common-subsequence

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')}");
}

Output:

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

EasyLang

Translation of: BASIC256

func$ right a$ n .
   return substr a$ (len a$ - n + 1) n
.
func$ left a$ n .
   if n < 0
      n = len a$ + n
   .
   return substr a$ 1 n
.
func$ lcs a$ b$ .
   if len a$ = 0 or len b$ = 0
      return ""
   .
   if right a$ 1 = right b$ 1
      return lcs left a$ -1 left b$ -1 & right a$ 1
   .
   x$ = lcs a$ left b$ -1
   y$ = lcs left a$ -1 b$
   if len x$ > len y$
      return x$
   else
      return y$
   .
.
print lcs "1234" "1224533324"
print lcs "thisisatest" "testing123testing"

Output:

1234
tsitest

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 version 1.3

Simple recursion

This solution is Brute force. It is slow

Translation of: 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

Translation of: 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")

Output:

tsitest
1234

Referring to LCS 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"

F#

Copied and slightly adapted from OCaml (direct recursion)

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

Output:

tsitest

Fortran

Works with: Fortran version 95

Using the iso_varying_string module which can be found 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

FreeBASIC

Function LCS(a As String, b As String) As String
    Dim As String x, y
    If Len(a) = 0 Or Len(b) = 0 Then 
        Return ""
    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 Return x Else Return y
    End If
End Function

Print LCS("1234", "1224533324")
Print LCS("thisisatest", "testing123testing")
Sleep

Go

Translation of: 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"));

Output:

1234
tsitest

Haskell

The Wikipedia solution translates directly into Haskell, with the only difference that equal characters are added in front:

longest xs ys = if length xs > 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"

Output:

"tsitest"

Icon and 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.

Library: Icon Programming Library

Uses deletec from strings

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

link strings

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

Output:

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 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

Translation of: 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

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

const lcs = (xx, yy) => {
  if (!xx.length || !yy.length) { return ''; }
  
  const [x, ...xs] = xx;
  const [y, ...ys] = yy;

  return (x === y) ? (x + 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 version 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")

Output:

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))
}

Output:

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

Logo

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/Wolfram Language

A built-in function can do this for us:

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

gives:

tsitest

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 LongestCommonSequence performs the puzzle-description.

Nim

Recursion

Translation of: Python

proc lcs(x, y: string): 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")

This recursive version is not efficient but the execution time can be greatly improved by using memoization.

Dynamic Programming

Translation of: Python

proc lcs(a, b: string): 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

from Haskell

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)

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
        Hashtbl.add cache (xs, ys) result;
        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

Translation of: Haskell

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

Translation of: 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.

Output:

:> ./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";

Alternate letting regex do all the work

use strict;
use warnings;
use feature 'bitwise';

print "lcs is ", lcs('thisisatest', 'testing123testing'), "\n";

sub lcs
  {
  my ($c, $d) = @_;
  for my $len ( reverse 1 .. length($c &. $d) )
    {
    "$c\n$d" =~ join '.*', ('(.)') x $len, "\n", map "\\$_", 1 .. $len and
      return join '', @{^CAPTURE};
    }
  return '';
  }

Output:

lcs is tastiest

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()).

with javascript_semantics
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

Output:

"1234"
"tsitest"

Alternate version

same output

with javascript_semantics
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

Picat

Wikipedia algorithm

With some added trickery for a 1-based language.

lcs_wiki(X,Y) = V => 
  [C, _Len] = lcs_length(X,Y),
  V = backTrace(C,X,Y,X.length+1,Y.length+1).

lcs_length(X, Y) = V=>
  M = X.length, 
  N = Y.length,
  C = [[0 : J in 1..N+1]  : I in 1..N+1],
  foreach(I in 2..M+1,J in 2..N+1)
     if X[I-1] == Y[J-1] then
        C[I,J] := C[I-1,J-1] + 1
     else
        C[I,J] := max([C[I,J-1], C[I-1,J]])
     end
  end,
  V = [C, C[M+1,N+1]].

backTrace(C, X, Y, I, J) = V =>
  if I == 1; J == 1 then
    V = ""
  elseif X[I-1] == Y[J-1] then
    V = backTrace(C, X, Y, I-1, J-1) ++ [X[I-1]]
  else 
    if C[I,J-1] > C[I-1,J] then
      V = backTrace(C, X, Y, I, J-1)
    else 
      V = backTrace(C, X, Y, I-1, J)
    end
  end.

Rule-based

Translation of: SETL

table
lcs_rule(A, B) = "", (A == ""; B == "") => true.
lcs_rule(A, B) = [A[1]] ++ lcs_rule(butfirst(A), butfirst(B)), A[1] == B[1] => true.
lcs_rule(A, B) = longest(lcs_rule(butfirst(A), B), lcs_rule(A, butfirst(B))) => true.

% Return the longest string of A and B
longest(A, B) = cond(A.length > B.length, A, B).
    
% butfirst (everything except first element)
butfirst(A) = [A[I] : I in 2..A.length].

Test

go =>
   Tests = [["thisisatest","testing123testing"],
            ["XMJYAUZ", "MZJAWXU"],
            ["1234", "1224533324"],
            ["beginning-middle-ending","beginning-diddle-dum-ending"]
            ],
   Funs = [lcs_wiki,lcs_rule],

   foreach(Fun in Funs)
     println(fun=Fun),
     foreach(Test in Tests)
        printf("%w : %w\n", Test, apply(Fun,Test[1],Test[2]))
     end,
     nl
   end,

   nl.

Output:

fun = lcs_wiki
[thisisatest,testing123testing] : tsitest
[XMJYAUZ,MZJAWXU] : MJAU
[1234,1224533324] : 1234
[beginning-middle-ending,beginning-diddle-dum-ending] : beginning-iddle-ending

fun = lcs_rule
[thisisatest,testing123testing] : tsitest
[XMJYAUZ,MZJAWXU] : MJAU
[1234,1224533324] : 1234
[beginning-middle-ending,beginning-diddle-dum-ending] : beginning-iddle-ending

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") ) )

Output:

-> ("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 ""

Output:

tsitest

Returns an array of integers:

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

Output:

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

Output:

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).

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

Translation of: 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 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 str(lcs(xs, ys)) + x
    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")

Output:

"tsitest">

Raku

(formerly Perl 6)

Recursion

Works with: rakudo version 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

Translation of: 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, $ystr)».comb;
    my (%positions, @Vs, $lcs);

    for @a.kv -> $i, $x { %positions{$x} +|= 1 +< ($i % @a) }

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

    my ($i, $j) = @a-1, @b-1;
    while ($i ≥ 0 and $j ≥ 0) {
        unless (@Vs[$j] +& (1 +< $i)) {
            $lcs [R~]= @a[$i] unless $j and ^@Vs[$j-1] +& (1 +< $i);
            $j--
        }
        $i--
    }
    $lcs
}

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

ReasonML

let longest = (xs, ys) =>
  if (List.length(xs) > List.length(ys)) {
    xs;
  } else {
    ys;
  };

let rec lcs = (a, b) =>
  switch (a, b) {
  | ([], _)
  | (_, []) => []
  | ([x, ...xs], [y, ...ys]) =>
    if (x == y) {
      [x, ...lcs(xs, ys)];
    } else {
      longest(lcs(a, ys), lcs(xs, b));
    }
  };

REXX

/*REXX program tests the  LCS  (Longest Common Subsequence)  subroutine.                */
parse arg aaa bbb .                              /*obtain optional arguments from the CL*/
say 'string A ='     aaa                         /*echo the string   A   to the screen. */
say 'string B ='     bbb                         /*  "   "     "     B    "  "     "    */
say '     LCS ='     LCS(aaa, bbb)               /*tell the  Longest Common Sequence.   */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
LCS: procedure; parse arg a,b,z                  /*Longest Common Subsequence.          */
                                                 /*reduce recursions, removes the       */
                                                 /*chars in  A ¬ in B,   and vice─versa.*/
     if z==''  then return LCS( LCS(a,b,0), LCS(b,a,0), 9)   /*Is Z null?   Do recurse. */
     j= length(a)
     if z==0 then do                             /*a 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 ''              /*Is 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

output when using the input of: 1234 1224533324

string A = 1234
string B = 1224533324
     LCS = 1234

output when using the input of: 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:

Ruby

Recursion

This solution is similar to the Haskell one. It is slow (The time complexity is exponential.)

Works with: Ruby version 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 version 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

Output:

tsitest
rsetod

Referring to LCS here and 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)
    su.intersect(sv).headOption match{
      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
  }

Output:

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
      }
    }
  }

Output:

 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

Translation of: C#

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

import <Utilities/Sequence.sl>;
    
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");

Output:

"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.first(1), xstr.slice(1),
                         ystr.first(1), ystr.slice(1))

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

say lcs("thisisatest", "testing123testing")

Output:

tsitest

Slate

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

Recursion

Translation of: 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

Translation of: 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

Swift 5.1

Recursion

rlcs(_ s1: String, _ s2: String) -> String {
   if s1.count == 0 || s2.count == 0 {
       return ""
   } else if s1[s1.index(s1.endIndex, offsetBy: -1)] == s2[s2.index(s2.endIndex, offsetBy: -1)] {
       return rlcs(String(s1[s1.startIndex..<s1.index(s1.endIndex, offsetBy: -1)]),
                   String(s2[s2.startIndex..<s2.index(s2.endIndex, offsetBy: -1)])) + String(s1[s1.index(s1.endIndex, offsetBy: -1)])
   } else {
       let str1 = rlcs(s1, String(s2[s2.startIndex..<s2.index(s2.endIndex, offsetBy: -1)]))
       let str2 = rlcs(String(s1[s1.startIndex..<s1.index(s1.endIndex, offsetBy: -1)]), s2)

       return str1.count > str2.count ? str1 : str2
   }
}

Dynamic Programming

func lcs(_ s1: String, _ s2: String) -> String {
    var lens = Array(
        repeating:Array(repeating: 0, count: s2.count + 1),
        count: s1.count + 1
    )
    
    for i in 0..<s1.count {
        for j in 0..<s2.count {
            if s1[s1.index(s1.startIndex, offsetBy: i)] == s2[s2.index(s2.startIndex, offsetBy: j)] {
                lens[i + 1][j + 1] = lens[i][j] + 1
            } else {
                lens[i + 1][j + 1] = max(lens[i + 1][j], lens[i][j + 1])
            }
        }
    }
    
    var returnStr = ""
    var x = s1.count
    var y = s2.count
    while x != 0 && y != 0 {
        if lens[x][y] == lens[x - 1][y] {
            x -= 1
        } else if lens[x][y] == lens[x][y - 1] {
            y -= 1
        } else {
            returnStr += String(s1[s1.index(s1.startIndex, offsetBy:  x - 1)])
            x -= 1
            y -= 1
        }
    }
    
    return String(returnStr.reversed())
}

Tcl

Recursive

Translation of: 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

Translation of: Java

Works with: Tcl version 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 && $y > 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')

Output:

'tsitest'

Wren

Translation of: Kotlin

var lcs // recursive
lcs = Fn.new { |x, y|
    if (x.count == 0 || y.count == 0) return ""
    var x1 = x[0...-1]
    var y1 = y[0...-1]
    if (x[-1] == y[-1]) return lcs.call(x1, y1) + x[-1]
    var x2 = lcs.call(x, y1)
    var y2 = lcs.call(x1, y)
    return (x2.count > y2.count) ? x2 : y2
}

var x = "thisisatest"
var y = "testing123testing"
System.print(lcs.call(x, y))

Output:

tsitest

zkl

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

Translation of: 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))

Output:

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