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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // JavaScript implementation of the approach let mod = 1e9 + 7; // Function to return the pairs of subsequences // from S[] and subsequences from T[] such // that both have the same content function subsequence(S, T, n, m) { // Create dp array let dp = new Array() for (let i = 0; i < n + 1; i++) { let temp = []; for (let j = 0; j < m + 1; j++) { temp.push([]) } dp.push(temp) } // Base values for (let i = 0; i <= n; i++) dp[i][0] = 1; // Base values for (let j = 0; j <= m; j++) dp[0][j] = 1; for (let i = 1; i <= n; ++i) { for (let 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 let S = [1, 1]; let n = S.length; let T = [1, 1]; let m = T.length; document.write(subsequence(S, T, n, m)); </script> |
6
Time Complexity: O( N*M )
Auxiliary Space: O(N*M )
Please Login to comment...