# Count of ways to generate a Matrix with product of each row and column as 1 or -1

Given two integers N and M, the task is to find the numbers of ways to form a matrix of size N * M consisting only of 1 or -1, such that the product of integers in each row and each column is equal to 1 or -1.
Examples:

Input: N = 2, M = 2
Output:
Explanation:
Possible ways to get product of each row and coloumn as 1 are,
{{1, 1}, {1, 1}} and {{-1, -1}, {-1, -1}}
Possible ways to get product of each row and coloumn as -1 are,
{{1, -1}, {-1, 1}} and {{-1, 1}, {1, -1}}
Hence, number of ways = 2 + 2 = 4

Input: N = 3, M = 3
Output: 32
Explanation:
There are 16 ways to get product as 1 and 16 ways to get product as -1.
Hence, number of ways = 16 + 16 = 32

Naive Approach:
The simplest approach to solve this problem is to generate all possible matrix of size N * M and for each of them, calculate the product of all rows and columns and check if it is 1 or -1.

Time complexity: O(2N*M)
Auxiliary Space: O(M*N)

Efficient Approach:
Assume, first N-1 rows and first M-1 columns are filled by 1 or -1. Now, the product of each row up to N-1 rows and each column up to M-1 columns would either be 1 or -1. There are a total 2 (N-1) * (M-1) Ways to form a matrix of size (N-1)*(M-1) filled with 1 or -1. Depending on what is needed as a product of N rows and M columns, the last row and column can be filled accordingly.
Follow the steps to solve the problem:

• If N + M is even
Number of possible matrices to get the product as 1 = 2 (N-1) * (M-1)
Number of possible matrices to get product as -1 = 2 (N-1) * (M-1)
• If N + M is odd
Number of possible matrices to get the product as 1 = 2 (N-1) * (M-1)
Number of possible matrices to get the product as -1 = 0

Below is the implementation of the above approach:

## C++

 `// C++ implementation of ` `// the above approach ` ` `  `#include  ` `using` `namespace` `std; ` ` `  `// Function to return the ` `// number of possible ways ` `void` `Solve(``int` `N, ``int` `M) ` `{ ` ` `  `    ``int` `temp = (N - 1) * (M - 1); ` `    ``int` `ans = ``pow``(2, temp); ` ` `  `    ``// Check if product can be -1 ` `    ``if` `((N + M) % 2 != 0) ` `        ``cout << ans; ` `    ``else` `        ``cout << 2 * ans; ` ` `  `    ``cout << endl; ` `} ` `// Driver Code ` `int` `main() ` `{ ` `    ``int` `N = 3; ` `    ``int` `M = 3; ` ` `  `    ``Solve(N, M); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `import` `java.util.Arrays; ` ` `  `class` `GFG{  ` `     `  `// Function to return the ` `// number of possible ways ` `static` `void` `Solve(``int` `N, ``int` `M) ` `{ ` `    ``int` `temp = (N - ``1``) * (M - ``1``); ` `    ``int` `ans = (``int``)(Math.pow(``2``, temp)); ` ` `  `    ``// Check if product can be -1 ` `    ``if` `((N + M) % ``2` `!= ``0``) ` `        ``System.out.print(ans); ` `    ``else` `        ``System.out.print(``2` `* ans); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{  ` `    ``int` `N = ``3``; ` `    ``int` `M = ``3``; ` `     `  `    ``Solve(N, M); ` `}  ` `}  ` ` `  `// This code is contributed by Shubham Prakash  `

## Python3

 `# Python3 program to implement ` `# the above approach ` `# Function to return  ` `# possible number of ways ` `def` `Solve(N, M): ` `  ``temp ``=` `(N ``-` `1``) ``*` `(M ``-` `1``) ` `  ``ans ``=` `pow``(``2``, temp) ` ` `  `  ``# Check if product can be -1 ` `  ``if` `((N ``+` `M) ``%` `2` `!``=` `0``): ` `    ``print``(ans) ` `    ``else``: ` `      ``print``(``2` `*` `ans) ` ` `  `      ``# driver code ` `      ``if` `__name__ ``=``=` `'__main__'``: ` `        ``N, M ``=` `3``, ``3` `        ``Solve(N, M) ` ` `  `# This code is contributed by Sri_srajit`

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to return the ` `// number of possible ways ` `static` `void` `Solve(``int` `N, ``int` `M) ` `{ ` `    ``int` `temp = (N - 1) * (M - 1); ` `    ``int` `ans = (``int``)(Math.Pow(2, temp)); ` ` `  `    ``// Check if product can be -1 ` `    ``if` `((N + M) % 2 != 0) ` `        ``Console.Write(ans); ` `    ``else` `        ``Console.Write(2 * ans); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``int` `N = 3; ` `    ``int` `M = 3; ` `     `  `    ``Solve(N, M); ` `} ` `} ` ` `  `// This code is contributed by rutvik_56 `

Output:

```32
```

Time complexity: O(log(N*M))
Auxiliary Space: O(1)

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