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++ 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 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 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# 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 |
<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 )
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size M+1.
- Set a base case by initializing the value of DP ( dp[0] = 1 ).
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create a variable prev used to store the previous value and temp to store current valueof DP vector .
- After every iteration assign the value of temp to prev for further iteration.
- At last return and print the final answer stored in dp[m].
Implementation:
// C++ code for above 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 vector
vector< int > dp(m + 1);
// Base values
dp[0] = 1;
for ( int i = 1; i <= n; ++i) {
int prev = 1;
for ( int j = 1; j <= m; ++j) {
int temp = dp[j];
dp[j] = dp[j] + dp[j - 1] - prev;
if (S[i - 1] == T[j - 1]) {
dp[j] = dp[j] + prev;
}
dp[j] += mod;
dp[j] %= mod;
prev = temp;
}
}
// Return the required answer
return dp[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) * 2 % mod;
return 0;
} |
import java.util.Arrays;
public class GFG {
static final int mod = ( int ) 1e9 + 7 ;
// 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 to store the number of ways to form subsequences from T
int [] dp = new int [m + 1 ];
// Base values
dp[ 0 ] = 1 ; // There is always an empty subsequence in any sequence
for ( int i = 1 ; i <= n; i++) {
int prev = 1 ; // Store the previous value of dp[j]
for ( int j = 1 ; j <= m; j++) {
int temp = dp[j]; // Store the current value of dp[j] to use in the next iteration
dp[j] = dp[j] + dp[j - 1 ] - prev;
if (S[i - 1 ] == T[j - 1 ]) {
dp[j] = dp[j] + prev; // If the elements match, add the previous value of dp[j]
}
dp[j] += mod; // Ensure dp[j] is non-negative
dp[j] %= mod; // Take the modulo to prevent overflow
prev = temp; // Update prev for the next iteration
}
}
// Return the required answer
return dp[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;
// Print the result
System.out.println(subsequence(S, T, n, m) * 2 % mod);
}
} |
# Python code for above approach # 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):
mod = ( 10 * * 9 ) + 7
# Create dp list
dp = [ 0 ] * (m + 1 )
# Base values
dp[ 0 ] = 1
for i in range ( 1 , n + 1 ):
prev = 1
for j in range ( 1 , m + 1 ):
temp = dp[j]
dp[j] = (dp[j] + dp[j - 1 ] - prev) % mod
if S[i - 1 ] = = T[j - 1 ]:
dp[j] = (dp[j] + prev) % mod
prev = temp
# Return the required answer
return dp[m]
# Driver code S = [ 1 , 1 ]
n = len (S)
T = [ 1 , 1 ]
m = len (T)
print ((subsequence(S, T, n, m) * 2 ) % ( 10 * * 9 + 7 ))
# This code is contributed by Susobhan Akhuli |
using System;
public class GFG
{ const int mod = 1000000007; // Value of mod constant
// Function to return the pairs of subsequences
// from S[] and subsequences from T[] such
// that both have the same content
public static int Subsequence( int [] S, int [] T, int n, int m)
{
// Create dp array to store the number of valid pairs
int [] dp = new int [m + 1];
// Base values
dp[0] = 1;
// Dynamic Programming: Fill the dp array
for ( int i = 1; i <= n; ++i)
{
int prev = 1;
for ( int j = 1; j <= m; ++j)
{
// Store the current value of dp[j] in temp
int temp = dp[j];
// Calculate the new value of dp[j] using the recurrence relation
dp[j] = (dp[j] + dp[j - 1] - prev + mod) % mod;
// Check if the elements at positions i-1 in S and j-1 in T are equal
if (S[i - 1] == T[j - 1])
{
// If they are equal, add the previous value (prev) to dp[j]
dp[j] = (dp[j] + prev) % mod;
}
// Update prev for the next iteration
prev = temp;
}
}
// Return the required answer
return dp[m];
}
public static void Main( string [] args)
{
int [] S = { 1, 1 };
int n = S.Length;
int [] T = { 1, 1 };
int m = T.Length;
// Calculate the number of pairs of subsequences with the same content
// and print the result, multiplied by 2 and then taken modulo mod
Console.WriteLine(Subsequence(S, T, n, m) * 2 % mod);
}
} |
const 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
const dp = new Array(m + 1).fill(0);
// Base values
dp[0] = 1;
for (let i = 1; i <= n; ++i) {
let prev = 1;
for (let j = 1; j <= m; ++j) {
const temp = dp[j];
dp[j] = dp[j] + dp[j - 1] - prev;
if (S[i - 1] === T[j - 1]) {
dp[j] = dp[j] + prev;
}
dp[j] += mod;
dp[j] %= mod;
prev = temp;
}
}
// Return the required answer
return dp[m];
} // Driver code const S = [1, 1]; const n = S.length; const T = [1, 1]; const m = T.length; console.log((subsequence(S, T, n, m) * 2) % mod); |
6
Time Complexity: O(N*M)
Auxiliary Space: O(M)