# Java Program for Longest Common Subsequence

*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])

`/* 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 ` |

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**Output:**

Length of LCS is 4

Following is a tabulated implementation for the LCS problem.

`/* 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 ` |

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**Output:**

Length of LCS is 4

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

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