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 */#include <bits/stdc++.h> int max(int a, int b); /* 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; } /* Driver program to test above function */int main() { char X[] = "AGGTAB"; char Y[] = "GXTXAYB"; int m = strlen(X); int n = strlen(Y); printf("Length of LCS is %d\n", lcs(X, Y, m, n)); return 0; } |
Length of LCS is 4
Following is a tabulated implementation for the LCS problem.
/* Dynamic Programming C/C++ implementation of LCS problem */#include <bits/stdc++.h> int max(int a, int b); /* 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[m + 1][n + 1]; int i, j; /* 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 (i = 0; i <= m; i++) { for (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]); } } /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */ return L[m][n]; } /* Utility function to get max of 2 integers */int max(int a, int b) { return (a > b) ? a : b; } /* Driver program to test above function */int main() { char X[] = "AGGTAB"; char Y[] = "GXTXAYB"; int m = strlen(X); int n = strlen(Y); printf("Length of LCS is %d\n", lcs(X, Y, m, n)); return 0; } |
Length of LCS is 4
Please refer complete article on Dynamic Programming | Set 4 (Longest Common Subsequence) for more details!
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