Find the equal pairs of subsequence of S and subsequence of T

Given two arrays S[] and T[] of size N and M respectively. The task is to find the pairs of subsequences of S[] and subsequences of T[] which are the same in content. Answer could be very large. So, print the answer modulo 109 + 7.

Examples:

Input: S[] = {1, 1}, T[] = {1, 1}
Output: 6
Subsequences of S[] are {}, {1}, {1} and {1, 1}.
Subsequences of T[] are {}, {1}, {1} and {1, 1}.
All the valid pairs are ({}, {}), ({1}, {1}), ({1}, {1}),
({1}, {1}), ({1}, {1}) and ({1, 1}, {1, 1}).



Input: S[] = {1, 3}, T[] = {3, 1}
Output: 3

Approach: Let dp[i][j] be the number of ways to create subsequences only using the first i elements of S[] and the first j elements of T[] such that the subsequences are the same and the ith element of S[] and the jth element of T[] are part of the subsequences.
Basically, dp[i][j] is the answer to the problem if only the first i elements of S[] and the first j elements of T[] are considered. If S[i] != T[j] then dp[i][j] = 0 because no subsequence will end by using the ith element of S[] and the jth element of T[]. If S[i] = T[j] then dp[i][j] = ∑k=1i-1 ∑l=1j-1 dp[k][l] + 1 because the previous index of S[] can be any index ≤ i and the previous index of T[] can be any index ≤ j.
As a base case, dp[0][0] = 1. This represents the case where no element is taken. The runtime of this is O(N2 * M2) but we can speed this up by precomputing the sums.
Let sum[i][j] = ∑k=1i ∑l=1jdp[k][l] which is a 2D prefix sum of the dp array. sum[i][j] = sum[i – 1][j] + sum[i][j – 1] – sum[i – 1][j – 1] + dp[i][j]. With sum[i][j], each state dp[i][j] can now be calculated in O(1).
Since there are N * M states, the runtime will be O(N * M).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define mod (int)(1e9 + 7)
  
// Function to return the pairs of subsequences
// from S[] and subsequences from T[] such
// that both have the same content
int subsequence(int S[], int T[], int n, int m)
{
  
    // Create dp array
    int dp[n + 1][m + 1];
  
    // Base values
    for (int i = 0; i <= n; i++)
        dp[i][0] = 1;
  
    // Base values
    for (int j = 0; j <= m; j++)
        dp[0][j] = 1;
  
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
  
            // Keep previous dp value
            dp[i][j] = dp[i - 1][j]
                       + dp[i][j - 1]
                       - dp[i - 1][j - 1];
  
            // If both elements are same
            if (S[i - 1] == T[j - 1])
                dp[i][j] += dp[i - 1][j - 1];
  
            dp[i][j] += mod;
            dp[i][j] %= mod;
        }
    }
  
    // Return the required answer
    return dp[n][m];
}
  
// Driver code
int main()
{
    int S[] = { 1, 1 };
    int n = sizeof(S) / sizeof(S[0]);
  
    int T[] = { 1, 1 };
    int m = sizeof(T) / sizeof(T[0]);
  
    cout << subsequence(S, T, n, m);
  
    return 0;
}

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Java














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// Java implementation of the approach  


import java.util.*;  


  


class GFG  


{  


  


// Function to return the pairs of subsequences  


// from S[] and subsequences from T[] such  


// that both have the same content  


static int subsequence(int[] S, int[] T,  


                       int n, int m)  


{  


  


    // Create dp array  


    int [][] dp = new int[n + 1][m + 1]; 


    int mod = 1000000007; 


  


    // Base values  


    for (int i = 0; i <= n; i++)  


        dp[i][0] = 1;  


  


    // Base values  


    for (int j = 0; j <= m; j++)  


        dp[0][j] = 1;  


  


    for (int i = 1; i <= n; ++i)  


    {  


        for (int j = 1; j <= m; ++j) 


        {  


  


            // Keep previous dp value  


            dp[i][j] = dp[i - 1][j] +  


                       dp[i][j - 1] -  


                       dp[i - 1][j - 1];  


  


            // If both elements are same  


            if (S[i - 1] == T[j - 1])  


                dp[i][j] += dp[i - 1][j - 1];  


  


            dp[i][j] += mod;  


            dp[i][j] %= mod;  


        }  


    }  


  


    // Return the required answer  


    return dp[n][m];  


}  


  


  


// Driver code  


public static void main(String []args)  


{  


    int S[] = { 1, 1 };  


    int n = S.length; 


  


    int T[] = { 1, 1 };  


    int m = T.length;  


  


    System.out.println(subsequence(S, T, n, m));  


}  


}  


  


// This code is contributed by Sanjit Prasad 



















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Python3














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# Python3 implementation of the approach  


import numpy as np 


  


mod = int(1e9 + 7)  


  


# Function to return the pairs of subsequences  


# from S[] and subsequences from T[] such  


# that both have the same content  


def subsequence(S, T, n, m) :  


  


    # Create dp array  


    dp = np.zeros((n + 1, m + 1));  


  


    # Base values  


    for i in range(n + 1) : 


        dp[i][0] = 1;  


  


    # Base values  


    for j in range(m + 1) :  


        dp[0][j] = 1;  


  


    for i in range(1, n + 1) : 


        for j in range(1, m + 1) : 


  


            # Keep previous dp value  


            dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - \ 


                       dp[i - 1][j - 1];  


  


            # If both elements are same  


            if (S[i - 1] == T[j - 1]) : 


                dp[i][j] += dp[i - 1][j - 1];  


  


            dp[i][j] += mod;  


            dp[i][j] %= mod;  


  


    # Return the required answer  


    return dp[n][m];  


  


# Driver code  


if __name__ == "__main__" :  


  


    S = [ 1, 1 ];  


    n = len(S);  


  


    T = [ 1, 1 ];  


    m = len(T);  


  


    print(subsequence(S, T, n, m));  


  


# This code is contributed by kanugargng 



















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C#














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// C# implementation of the approach  


using System; 


  


class GFG  


{  


  


    // Function to return the pairs of subsequences  


    // from S[] and subsequences from T[] such  


    // that both have the same content  


    static int subsequence(int[] S, int[] T,  


                           int n, int m)  


    {  


      


        // Create dp array  


        int [,] dp = new int[n + 1, m + 1];  


        int mod = 1000000007;  


      


        // Base values  


        for (int i = 0; i <= n; i++)  


            dp[i, 0] = 1;  


      


        // Base values  


        for (int j = 0; j <= m; j++)  


            dp[0, j] = 1;  


      


        for (int i = 1; i <= n; ++i)  


        {  


            for (int j = 1; j <= m; ++j)  


            {  


      


                // Keep previous dp value  


                dp[i, j] = dp[i - 1, j] +  


                           dp[i, j - 1] -  


                           dp[i - 1, j - 1];  


      


                // If both elements are same  


                if (S[i - 1] == T[j - 1])  


                    dp[i, j] += dp[i - 1, j - 1];  


      


                dp[i, j] += mod;  


                dp[i, j] %= mod;  


            }  


        }  


      


        // Return the required answer  


        return dp[n, m];  


    }  


      


    // Driver code  


    public static void Main()  


    {  


        int []S = { 1, 1 };  


        int n = S.Length;  


      


        int []T = { 1, 1 };  


        int m = T.Length;  


      


        Console.WriteLine(subsequence(S, T, n, m));  


    }  


}  


  


// This code is contributed by AnkitRai01 



















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Output:


6






          
          
          

          

    

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