Find length of longest subsequence of one string which is substring of another string

Given two string X and Y. The task is to find the length of longest subsequence of string X which is substring in sequence Y.

Examples:

Input : X = "ABCD",  Y = "BACDBDCD"
Output : 3
"ACD" is longest subsequence of X which
is substring of Y.

Input : X = "A",  Y = "A"
Output : 1

Method 1 (Brute Force):
Use brute force to find all the subsequence of X and for each subsequence check whether it is substring of Y or not. If it is substring of Y, maintain a maximum length varible and compare length with it.

Method 2: (Dynamic Programming):
Let n be length of X and m be length of Y. Create a 2D array ‘dp[][]’ of m + 1 rows and n + 1 columns. Value dp[i][j] is maximum length of subsequence of X[0….j] which is substring of Y[0….i]. Now for each cell of dp[][] fill value as :

for (i = 1 to m)
  for (j = 1 to n)
    if (x[i-1] == y[j - 1])
      dp[i][j] = dp[i-1][j-1] + 1;
    else
      dp[i][j] = dp[i][j-1];

And finally, the length of the longest subsequence of x which is substring of y is max(dp[i][n]) where 1 <= i <= m.

Below is implementation this approach:

C/C++

// C++ program to find maximum length of
// subsequence of a string X such it is
// substring in another string Y.
#include <bits/stdc++.h>
#define MAX 1000
using namespace std;

// Return the maximum size of substring of
// X which is substring in Y.
int maxSubsequenceSubstring(char x[], char y[],
                            int n, int m)
{
    int dp[MAX][MAX];

    // Initialize the dp[][] to 0.
    for (int i = 0; i <= m; i++)
        for (int j = 0; j <= n; j++)
            dp[i][j] = 0;

    // Calculating value for each element.
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {

            // If alphabet of string X and Y are
            // equal make dp[i][j] = 1 + dp[i-1][j-1]
            if (x[j - 1] == y[i - 1])
                dp[i][j] = 1 + dp[i - 1][j - 1];

            // Else copy the previous value in the
            // row i.e dp[i-1][j-1]
            else
                dp[i][j] = dp[i][j - 1];
        }
    }

    // Finding the maximum length.
    int ans = 0;
    for (int i = 1; i <= m; i++)
        ans = max(ans, dp[i][n]);

    return ans;
}

// Driver Program
int main()
{
    char x[] = "ABCD";
    char y[] = "BACDBDCD";
    int n = strlen(x), m = strlen(y);
    cout << maxSubsequenceSubstring(x, y, n, m);
    return 0;
}

Java

// Java program to find maximum length of
// subsequence of a string X such it is
// substring in another string Y.

public class GFG 
{
	static final int MAX = 1000;
	
	// Return the maximum size of substring of
	// X which is substring in Y.
	static int maxSubsequenceSubstring(char x[], char y[],
	                            int n, int m)
	{
	    int dp[][] = new int[MAX][MAX];
	 
	    // Initialize the dp[][] to 0.
	    for (int i = 0; i <= m; i++)
	        for (int j = 0; j <= n; j++)
	            dp[i][j] = 0;
	 
	    // Calculating value for each element.
	    for (int i = 1; i <= m; i++) {
	        for (int j = 1; j <= n; j++) {
	 
	            // If alphabet of string X and Y are
	            // equal make dp[i][j] = 1 + dp[i-1][j-1]
	            if (x[j - 1] == y[i - 1])
	                dp[i][j] = 1 + dp[i - 1][j - 1];
	 
	            // Else copy the previous value in the
	            // row i.e dp[i-1][j-1]
	            else
	                dp[i][j] = dp[i][j - 1];
	        }
	    }
	 
	    // Finding the maximum length.
	    int ans = 0;
	    for (int i = 1; i <= m; i++)
	        ans = Math.max(ans, dp[i][n]);
	 
	    return ans;
	}
	
	// Driver Method
	public static void main(String[] args)
	{
		char x[] = "ABCD".toCharArray();
	    char y[] = "BACDBDCD".toCharArray();
	    int n = x.length, m = y.length;
	    System.out.println(maxSubsequenceSubstring(x, y, n, m));
	}
}


Output:

3

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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