# Count ways to select N pairs of candies of distinct colors (Dynamic Programming + Bitmasking)

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2022

Given an integer N representing the number of red and blue candies and a matrix mat[][] of size N * N, where mat[i][j] = 1 represents the existence of a pair between ith red candy and jth blue candy, the task is to find the count of ways to select N pairs of candies such that each pair contains distinct candies of different colors.

Examples:

Input: N = 2, mat[][] = { { 1, 1 }, { 1, 1 } }
Output:
Explanation:
Possible ways to select N (= 2) pairs of candies are { { (1, 1), (2, 2) }, { (1, 2), (2, 1) } }.
Therefore, the required output is 2.

Input: N = 3, mat[][] = { { 0, 1, 1 }, { 1, 0, 1 }, { 1, 1, 1 } }
Output:
Explanation:
Possible ways to select N (= 3) pairs of candies are: { { (1, 2), (2, 1), (3, 3) }, { (1, 2), (2, 3), (3, 1) }, { (1, 3), (2, 1), (3, 2) } }
Therefore, the required output is 2.

Naive Approach: The simplest approach to solve this problem is to generate all possible permutations of N pairs containing distinct candies of different colors. Finally, print the count obtained.

Below is the implementation of the above approach:

## C++14

 `// C++14 program to implement ` `// the above approach ` `#include ` `using` `namespace` `std;`   `// Function to count ways to select N distinct` `// pairs of candies with different colours` `int` `numOfWays(vector> a, ``int` `n,` `                 ``int` `i, set<``int``> &blue)` `{` `    `  `    ``// If n pairs are selected` `    ``if` `(i == n)` `        ``return` `1;`   `    ``// Stores count of ways` `    ``// to select the i-th pair` `    ``int` `count = 0;`   `    ``// Iterate over the range [0, n]` `    ``for``(``int` `j = 0; j < n; j++)` `    ``{` `        `  `        ``// If pair (i, j) is not included` `        ``if` `(a[i][j] == 1 && blue.find(j) == blue.end())` `        ``{` `            ``blue.insert(j);` `            ``count += numOfWays(a, n, i + 1, blue);` `            ``blue.erase(j);` `        ``}` `    ``}` `    ``return` `count;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 3;` `    ``vector> mat = { { 0, 1, 1 },` `                                ``{ 1, 0, 1 },` `                                ``{ 1, 1, 1 } };` `    ``set<``int``> mpp;` `    `  `    ``cout << (numOfWays(mat, n, 0, mpp));` `}`   `// This code is contributed by mohit kumar 29`

## Java

 `// Java program to implement ` `// the above approach ` `import` `java.io.*;` `import` `java.lang.*;` `import` `java.util.*;`   `class` `GFG{`   `// Function to count ways to select N distinct` `// pairs of candies with different colours` `static` `int` `numOfWays(``int` `a[][], ``int` `n, ``int` `i,` `                     ``HashSet blue)` `{` `    `  `    ``// If n pairs are selected` `    ``if` `(i == n)` `        ``return` `1``;`   `    ``// Stores count of ways` `    ``// to select the i-th pair` `    ``int` `count = ``0``;`   `    ``// Iterate over the range [0, n]` `    ``for``(``int` `j = ``0``; j < n; j++) ` `    ``{` `        `  `        ``// If pair (i, j) is not included` `        ``if` `(a[i][j] == ``1` `&& !blue.contains(j)) ` `        ``{` `            ``blue.add(j);` `            ``count += numOfWays(a, n, i + ``1``, blue);` `            ``blue.remove(j);` `        ``}` `    ``}` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``3``;` `    ``int` `mat[][] = { { ``0``, ``1``, ``1` `}, ` `                    ``{ ``1``, ``0``, ``1` `}, ` `                    ``{ ``1``, ``1``, ``1` `} };` `    ``HashSet mpp = ``new` `HashSet<>();`   `    ``System.out.println((numOfWays(mat, n, ``0``, mpp)));` `}` `}`   `// This code is contributed by Kingash`

## Python3

 `# Python3 program to implement` `# the above approach`     `# Function to count ways to select N distinct ` `# pairs of candies with different colours` `def` `numOfWays(a, n, i ``=` `0``, blue ``=` `[]):`   `    ``# If n pairs are selected` `    ``if` `i ``=``=` `n:` `        ``return` `1`   `    ``# Stores count of ways ` `    ``# to select the i-th pair` `    ``count ``=` `0`   `    ``# Iterate over the range [0, n]` `    ``for` `j ``in` `range``(n):`   `        ``# If pair (i, j) is not included` `        ``if` `mat[i][j] ``=``=` `1` `and` `j ``not` `in` `blue:` `            ``count ``+``=` `numOfWays(mat, n, i ``+` `1``, ` `                                ``blue ``+` `[j])` `                                `  `    ``return` `count`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `3` `    ``mat ``=` `[ [``0``, ``1``, ``1``],` `            ``[``1``, ``0``, ``1``], ` `            ``[``1``, ``1``, ``1``] ]` `    ``print``(numOfWays(mat, n))`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{` `    `  `// Function to count ways to select N distinct` `// pairs of candies with different colours` `static` `int` `numOfWays(``int``[,] a, ``int` `n, ``int` `i,` `                     ``HashSet<``int``> blue)` `{` `    `  `    ``// If n pairs are selected` `    ``if` `(i == n)` `        ``return` `1;`   `    ``// Stores count of ways` `    ``// to select the i-th pair` `    ``int` `count = 0;`   `    ``// Iterate over the range [0, n]` `    ``for``(``int` `j = 0; j < n; j++)` `    ``{` `        `  `        ``// If pair (i, j) is not included` `        ``if` `(a[i, j] == 1 && !blue.Contains(j))` `        ``{` `            ``blue.Add(j);` `            ``count += numOfWays(a, n, i + 1, blue);` `            ``blue.Remove(j);` `        ``}` `    ``}` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    ``int` `n = 3;` `    ``int``[,] mat = { { 0, 1, 1 },` `                   ``{ 1, 0, 1 }, ` `                   ``{ 1, 1, 1 } };` `    ``HashSet<``int``> mpp = ``new` `HashSet<``int``>();`   `    ``Console.WriteLine((numOfWays(mat, n, 0, mpp)));` `}` `}`   `// This code is contributed by ukasp`

## Javascript

 ``

Output:

`3`

Time complexity: O(N!)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized for Dynamic programming with Bit masking. Instead of generating all permutations of N blue candies, for every red candy, use a mask, where jth bit of mask checks if jth blue candy is available for selecting the current pair or not.
The recurrence relation to solving the problem is as follows:

If Kth bit of mask is unset and mat[i][k] = 1:
dp[i + 1][j | (1 << k)] += dp[i][j]

where, (j | (1 << k)) marks the kth blue candy as selected.
dp[i][j] = Count of ways to make pairs between i red candy and N blue candies, where j is a permutation of N bit number ranging from 0 to 2N – 1).

Follow the steps below to solve the problem:

• Initialize a 2D array, say dp[][], where dp[i][j] stores the count of ways to make pairs between i red candies and N blue candies. j represents a permutation of N bit number ranging from 0 to 2N-1.
• Use the above recurrence relation and fill all possible dp states of the recurrence relation.
• Finally, print the dp state where there are N red candies and N blue candies are selected, i.e. dp[i][2n-1].

Below is the implementation of the above approach:

## Python3

 `# Python program to implement` `# the above approach` ` `  ` `  `# Function to count ways to select N distinct ` `# pairs of candies with different colors` `def` `numOfWays(a, n):`     `    ``# dp[i][j]: Stores count of ways to make` `    ``# pairs between i red candies and N blue candies` `    ``dp ``=` `[[``0``]``*``((``1` `<< n)``+``1``) ``for` `_ ``in` `range``(n ``+` `1``)]`   `    ``# If there is no red and blue candy,` `    ``# the count of ways to make n pairs is 1` `    ``dp[``0``][``0``] ``=` `1`   `    ``# i: Stores count of red candy` `    ``for` `i ``in` `range``(n):`   `        ``# j generates a permutation of blue` `        ``# candy of selected / not selected ` `        ``# as a binary number with n bits` `        ``for` `j ``in` `range``(``1` `<< n):`   `            ``if` `dp[i][j] ``=``=` `0``:` `                ``continue` `   `  `            ``# Iterate over the range [0, n]    ` `            ``for` `k ``in` `range``(n):`   `                ``# Create a mask with only` `                ``# the k-th bit as set` `                ``mask ``=` `1` `<< k` `                `  `                ``# If Kth bit of mask is` `                ``# unset  and mat[i][k] = 1` `                ``if` `not``(mask & j) ``and` `a[i][k]:` `                    ``dp[i ``+` `1``][j | mask] ``+``=` `dp[i][j]` `                    `  `        `  `                    `  `    ``# Return the dp states, where n red` `    ``# and n blue candies are selected` `    ``return` `dp[n][(``1` `<< n)``-``1``]`       `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `3` `    ``mat ``=` `[[``0``, ``1``, ``1``],` `           ``[``1``, ``0``, ``1``],` `           ``[``1``, ``1``, ``1``]]` `    ``print``(numOfWays(mat, n))`

## C#

 `// C# program to implement above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `  ``// Function to count ways to select N distinct ` `  ``// pairs of candies with different colors` `  ``static` `int` `numOfWays(``int``[][] a,``int` `n){`     `    ``// dp[i][j]: Stores count of ways to make` `    ``// pairs between i red candies and N blue candies` `    ``int``[][] dp = ``new` `int``[n+1][];` `    ``for``(``int` `i = 0 ; i <= n ; i++){` `      ``dp[i] = ``new` `int``[(1 << n)+1];` `    ``}`   `    ``// If there is no red and blue candy,` `    ``// the count of ways to make n pairs is 1` `    ``dp = 1;`   `    ``// i: Stores count of red candy` `    ``for``(``int` `i = 0 ; i < n ; i++){`   `      ``// j generates a permutation of blue` `      ``// candy of selected / not selected ` `      ``// as a binary number with n bits` `      ``for``(``int` `j = 0 ; j < (1 << n) ; j++){`   `        ``if` `(dp[i][j] == 0){` `          ``continue``;` `        ``}`   `        ``// Iterate over the range [0, n] ` `        ``for` `(``int` `k = 0 ; k < n ; k++){`   `          ``// Create a mask with only` `          ``// the k-th bit as set` `          ``int` `mask = (1 << k);`   `          ``// If Kth bit of mask is` `          ``// unset and mat[i][k] = 1` `          ``if` `((mask & j) == 0 && a[i][k] == 1){` `            ``dp[i + 1][j | mask] += dp[i][j];` `          ``}` `        ``}` `      ``}` `    ``}`   `    ``// Return the dp states, where n red` `    ``// and n blue candies are selected` `    ``return` `dp[n][(1 << n)-1];` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main(``string``[] args){`   `    ``int` `n = 3;` `    ``int``[][] mat = ``new` `int``[];` `    ``mat = ``new` `int``[]{0, 1, 1};` `    ``mat = ``new` `int``[]{1, 0, 1};` `    ``mat = ``new` `int``[]{1, 1, 1};` `    ``Console.Write(numOfWays(mat, n));`   `  ``}` `}`   `// This code is contributed by subhamgoyal2014.`

Output:

`3`

Time Complexity:O(N2 * 2N)
Auxiliary Space: O(N * 2N)

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