Java Program for Longest Common Subsequence

Last Updated : 23 Jul, 2025

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