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 |
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 |
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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