C++ 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])

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/* 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;
}
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Output:
Length of LCS is 4

Following is a tabulated implementation for the LCS problem.

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/* 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;
}
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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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