# Java Program for Longest Common Subsequence

Last Updated : 08 Mar, 2023

LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, “abc”, “abg”, “bdf”, “aeg”, ‘”acefg”, .. etc are subsequences of “abcdefg”. So a string of length n has 2^n different possible subsequences. It is a classic computer science problem, the basis of diff (a file comparison program that outputs the differences between two files), and has applications in bioinformatics. Examples: LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3. LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4. Let the input sequences be X[0..m-1] and Y[0..n-1] of lengths m and n respectively. And let L(X[0..m-1], Y[0..n-1]) be the length of LCS of the two sequences X and Y. Following is the recursive definition of L(X[0..m-1], Y[0..n-1]). If last characters of both sequences match (or X[m-1] == Y[n-1]) then L(X[0..m-1], Y[0..n-1]) = 1 + L(X[0..m-2], Y[0..n-2]) If last characters of both sequences do not match (or X[m-1] != Y[n-1]) then L(X[0..m-1], Y[0..n-1]) = MAX ( L(X[0..m-2], Y[0..n-1]), L(X[0..m-1], Y[0..n-2])

## Java

 `/* A Naive recursive implementation of LCS problem in java*/` `public` `class` `LongestCommonSubsequence {`   `    ``/* Returns length of LCS for X[0..m-1], Y[0..n-1] */` `    ``int` `lcs(``char``[] X, ``char``[] Y, ``int` `m, ``int` `n)` `    ``{` `        ``if` `(m == ``0` `|| n == ``0``)` `            ``return` `0``;` `        ``if` `(X[m - ``1``] == Y[n - ``1``])` `            ``return` `1` `+ lcs(X, Y, m - ``1``, n - ``1``);` `        ``else` `            ``return` `max(lcs(X, Y, m, n - ``1``), lcs(X, Y, m - ``1``, n));` `    ``}`   `    ``/* Utility function to get max of 2 integers */` `    ``int` `max(``int` `a, ``int` `b)` `    ``{` `        ``return` `(a > b) ? a : b;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``LongestCommonSubsequence lcs = ``new` `LongestCommonSubsequence();` `        ``String s1 = "AGGTAB";` `        ``String s2 = "GXTXAYB";`   `        ``char``[] X = s1.toCharArray();` `        ``char``[] Y = s2.toCharArray();` `        ``int` `m = X.length;` `        ``int` `n = Y.length;`   `        ``System.out.println("Length of LCS is"` `                           ``+ " " + lcs.lcs(X, Y, m, n));` `    ``}` `}`   `// This Code is Contributed by Saket Kumar`

Output:

`Length of LCS is 4`

Time Complexity: O(2^(N+M)) where N and M are the lengths of two input strings.

Space Complexity: O(N+M) for recursion stack used.

Following is a tabulated implementation for the LCS problem.

## Java

 `/* Dynamic Programming Java implementation of LCS problem */` `public` `class` `LongestCommonSubsequence {`   `    ``/* Returns length of LCS for X[0..m-1], Y[0..n-1] */` `    ``int` `lcs(``char``[] X, ``char``[] Y, ``int` `m, ``int` `n)` `    ``{` `        ``int` `L[][] = ``new` `int``[m + ``1``][n + ``1``];`   `        ``/* Following steps build L[m+1][n+1] in bottom up fashion. Note` `         ``that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */` `        ``for` `(``int` `i = ``0``; i <= m; i++) {` `            ``for` `(``int` `j = ``0``; j <= n; j++) {` `                ``if` `(i == ``0` `|| j == ``0``)` `                    ``L[i][j] = ``0``;` `                ``else` `if` `(X[i - ``1``] == Y[j - ``1``])` `                    ``L[i][j] = L[i - ``1``][j - ``1``] + ``1``;` `                ``else` `                    ``L[i][j] = max(L[i - ``1``][j], L[i][j - ``1``]);` `            ``}` `        ``}` `        ``return` `L[m][n];` `    ``}`   `    ``/* Utility function to get max of 2 integers */` `    ``int` `max(``int` `a, ``int` `b)` `    ``{` `        ``return` `(a > b) ? a : b;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``LongestCommonSubsequence lcs = ``new` `LongestCommonSubsequence();` `        ``String s1 = "AGGTAB";` `        ``String s2 = "GXTXAYB";`   `        ``char``[] X = s1.toCharArray();` `        ``char``[] Y = s2.toCharArray();` `        ``int` `m = X.length;` `        ``int` `n = Y.length;`   `        ``System.out.println("Length of LCS is"` `                           ``+ " " + lcs.lcs(X, Y, m, n));` `    ``}` `}`   `// This Code is Contributed by Saket Kumar`

Output:

`Length of LCS is 4`

Time Complexity: O(m*n) where m and n are lengths of two input strings.

Space Complexity: O(m*n) as 2d array has been created.

Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!

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