# 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 `

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 `

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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