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 variable 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++

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// 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;
}

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Java














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// 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)); 


    } 


} 



















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Python3














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# Python3 program to find maximum  


# length of subsequence of a string 


# X such it is substring in another 


# string Y.  


  


MAX = 1000


  


# Return the maximum size of  


# substring of X which is 


# substring in Y. 


def maxSubsequenceSubstring(x, y, n, m): 


    dp = [[0 for i in range(MAX)] 


             for i in range(MAX)] 


               


    # Initialize the dp[][] to 0. 


  


    # Calculating value for each element. 


    for i in range(1, m + 1): 


        for j in range(1, n + 1): 


              


            # 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 


    ans = 0


    for i in range(1, m + 1): 


        ans = max(ans, dp[i][n]) 


    return ans 


  


# Driver Code  


x = "ABCD"


y = "BACDBDCD"


n = len(x) 


m = len(y) 


print(maxSubsequenceSubstring(x, y, n, m)) 


  


# This code is contributed  


# by sahilshelangia 



















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C#














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// C# program to find maximum length of 


// subsequence of a string X such it is 


// substring in another string Y. 


using System; 


  


public class GFG  


{ 


    static int MAX = 1000; 


      


    // Return the maximum size of substring of 


    // X which is substring in Y. 


    static int maxSubsequenceSubstring(string x, string 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 x = "ABCD"; 


        string y = "BACDBDCD"; 


        int n = x.Length, m = y.Length; 


          


        Console.WriteLine(maxSubsequenceSubstring(x, 


                                            y, n, m)); 


    } 


} 


  


// This code is contributed by vt_m. 



















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PHP














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<?php 


// PHP program to find maximum length of 


// subsequence of a string X such it is 


// substring in another string Y. 


  


// Return the maximum size of substring of 


// X which is substring in Y. 


function maxSubsequenceSubstring($x, $y, 


                                 $n, $m) 


{ 


    $dp; 


  


    // Initialize the dp[][] to 0. 


    for ($i = 0; $i <= $m; $i++) 


        for ($j = 0; $j <= $n; $j++) 


            $dp[$i][$j] = 0; 


  


    // Calculating value for each element. 


    for ($i = 1; $i <= $m; $i++) { 


        for ( $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. 


    $ans = 0; 


    for ( $i = 1; $i <= $m; $i++) 


        $ans = max($ans, $dp[$i][$n]); 


  


    return $ans; 


} 


  


// Driver Code 


{ 


    $x = "ABCD"; 


    $y = "BACDBDCD"; 


    $n = strlen($x); $m = strlen($y); 


    echo maxSubsequenceSubstring($x, $y, $n, $m); 


    return 0; 


} 


  


// This code is contributed by nitin mittal 


?> 



















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Output:


3



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