# Ways of filling matrix such that product of all rows and all columns are equal to unity

We are given three values , and where is number of rows in matrix, is number of columns in the matrix and is the number that can have only two values -1 and 1. Our aim is to find the number of ways of filling the matrix of such that the product of all the elements in each row and each column is equal to . Since the number of ways can be large we will output
Examples:

```Input : n = 2, m = 4, k = -1
Output : 8
Following configurations satisfy the conditions:-

Input  : n = 2, m = 1, k = -1
Output : The number of filling the matrix
are 0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

From the above conditions, it is clear that the only elements that can be entered in the matrix are 1 and -1. Now we can easily deduce some of the corner cases

1. If k = -1, then the sum of number of rows and columns cannot be odd because -1 will be present odd
number of times in each row and column therefore if the sum is odd then answer is .
2. If n = 1 or m = 1 then there is only one way of filling the matrix therefore answer is 1.
3. If none of the above cases are applicable then we fill the first rows and the first columns with 1 and -1. Then the remaining numbers can be uniquely identified since the product of each row an each column is already known therefore the answer is .

## C++

 `// CPP program to find number of ways to fill ` `// a matrix under given constraints ` `#include ` `using` `namespace` `std; ` ` `  `#define mod 100000007 ` ` `  `/* Returns a raised power t under modulo mod */` `long` `long` `modPower(``long` `long` `a, ``long` `long` `t) ` `{ ` `    ``long` `long` `now = a, ret = 1; ` ` `  `    ``// Counting number of ways of filling the matrix ` `    ``while` `(t) { ` `        ``if` `(t & 1) ` `            ``ret = now * (ret % mod); ` `        ``now = now * (now % mod); ` `        ``t >>= 1; ` `    ``} ` `    ``return` `ret; ` `} ` ` `  `// Function calculating the answer ` `long` `countWays(``int` `n, ``int` `m, ``int` `k) ` `{ ` `    ``// if sum of numbers of rows and columns is odd ` `    ``// i.e (n + m) % 2 == 1 and k = -1 then there ` `    ``// are 0 ways of filiing the matrix. ` `    ``if` `(k == -1 && (n + m) % 2 == 1) ` `        ``return` `0; ` ` `  `    ``// If there is one row or one column then there ` `    ``// is only one way of filling the matrix ` `    ``if` `(n == 1 || m == 1) ` `        ``return` `1; ` ` `  `    ``// If the above cases are not followed then we ` `    ``// find ways to fill the n - 1 rows and m - 1 ` `    ``// columns which is 2 ^ ((m-1)*(n-1)). ` `    ``return` `(modPower(modPower((``long` `long``)2, n - 1), ` `                                    ``m - 1) % mod); ` `} ` ` `  `// Driver function for the program ` `int` `main() ` `{ ` `    ``int` `n = 2, m = 7, k = 1; ` `    ``cout << countWays(n, m, k); ` `    ``return` `0; ` `} `

Output:

```64
```

## Java

 `// Java program to find number of ways to fill ` `// a matrix under given constraints ` `import` `java.io.*; ` ` `  `class` `Example { ` ` `  `    ``final` `static` `long` `mod = ``100000007``; ` ` `  `    ``/* Returns a raised power t under modulo mod */` `    ``static` `long` `modPower(``long` `a, ``long` `t, ``long` `mod) ` `    ``{ ` `        ``long` `now = a, ret = ``1``; ` ` `  `        ``// Counting number of ways of filling the ` `        ``// matrix ` `        ``while` `(t > ``0``) { ` `            ``if` `(t % ``2` `== ``1``) ` `                ``ret = now * (ret % mod); ` `            ``now = now * (now % mod); ` `            ``t >>= ``1``; ` `        ``} ` `        ``return` `ret; ` `    ``} ` ` `  `    ``// Function calculating the answer ` `    ``static` `long` `countWays(``int` `n, ``int` `m, ``int` `k) ` `    ``{ ` `        ``// if sum of numbers of rows and columns is ` `        ``// odd i.e (n + m) % 2 == 1 and k = -1, ` `        ``// then there are 0 ways of filiing the matrix. ` `        ``if` `(n == ``1` `|| m == ``1``) ` `            ``return` `1``; ` ` `  `        ``// If there is one row or one column then ` `        ``// there is only one way of filling the matrix ` `        ``else` `if` `((n + m) % ``2` `== ``1` `&& k == -``1``) ` `            ``return` `0``; ` ` `  `       ``// If the above cases are not followed then we ` `       ``// find ways to fill the n - 1 rows and m - 1 ` `       ``// columns which is 2 ^ ((m-1)*(n-1)). ` `        ``return` `(modPower(modPower((``long``)``2``, n - ``1``, mod), ` `                                    ``m - ``1``, mod) % mod); ` `    ``} ` ` `  `    ``// Driver function for the program ` `    ``public` `static` `void` `main(String args[]) ``throws` `IOException ` `    ``{ ` `        ``int` `n = ``2``, m = ``7``, k = ``1``; ` `        ``System.out.println(countWays(n, m, k)); ` `    ``} ` `} `

## Python 3

 `# Python program to find number of ways to  ` `# fill a matrix under given constraints ` ` `  `# Returns a raised power t under modulo mod  ` `def` `modPower(a, t): ` `     `  `    ``now ``=` `a; ` `    ``ret ``=` `1``; ` `    ``mod ``=` `100000007``; ` ` `  `    ``# Counting number of ways of filling ` `    ``# the matrix ` `    ``while` `(t):  ` `        ``if` `(t & ``1``): ` `            ``ret ``=` `now ``*` `(ret ``%` `mod); ` `        ``now ``=` `now ``*` `(now ``%` `mod); ` `        ``t >>``=` `1``; ` `     `  `    ``return` `ret; ` ` `  `# Function calculating the answer ` `def` `countWays(n, m, k): ` ` `  `    ``mod``=` `100000007``; ` `     `  `    ``# if sum of numbers of rows and columns  ` `    ``# is odd i.e (n + m) % 2 == 1 and k = -1  ` `    ``# then there are 0 ways of filiing the matrix. ` `    ``if` `(k ``=``=` `-``1` `and` `((n ``+` `m) ``%` `2` `=``=` `1``)): ` `        ``return` `0``; ` ` `  `    ``# If there is one row or one column then  ` `    ``# there is only one way of filling the matrix ` `    ``if` `(n ``=``=` `1` `or` `m ``=``=` `1``): ` `        ``return` `1``; ` ` `  `    ``# If the above cases are not followed then we ` `    ``# find ways to fill the n - 1 rows and m - 1 ` `    ``# columns which is 2 ^ ((m-1)*(n-1)). ` `    ``return` `(modPower(modPower(``2``, n ``-` `1``),  ` `                              ``m ``-` `1``) ``%` `mod); ` ` `  `# Driver Code ` `n ``=` `2``; ` `m ``=` `7``; ` `k ``=` `1``; ` `print``(countWays(n, m, k)); ` ` `  `# This code is contributed  ` `# by Shivi_Aggarwal `

## C#

 `// C# program to find number of ways to fill ` `// a matrix under given constraints ` `using` `System; ` ` `  `class` `Example ` `{ ` ` `  `    ``static` `long` `mod = 100000007; ` ` `  `    ``// Returns a raised power t  ` `    ``// under modulo mod  ` `    ``static` `long` `modPower(``long` `a, ``long` `t, ` `                         ``long` `mod) ` `    ``{ ` `        ``long` `now = a, ret = 1; ` ` `  `        ``// Counting number of ways  ` `        ``// of filling the ` `        ``// matrix ` `        ``while` `(t > 0) ` `        ``{ ` `            ``if` `(t % 2 == 1) ` `                ``ret = now * (ret % mod); ` `            ``now = now * (now % mod); ` `            ``t >>= 1; ` `        ``} ` `        ``return` `ret; ` `    ``} ` ` `  `    ``// Function calculating the answer ` `    ``static` `long` `countWays(``int` `n, ``int` `m, ` `                          ``int` `k) ` `    ``{ ` `        ``// if sum of numbers of rows  ` `        ``// and columns is odd i.e ` `        ``// (n + m) % 2 == 1 and  ` `        ``// k = -1, then there are 0  ` `        ``// ways of filiing the matrix. ` `        ``if` `(n == 1 || m == 1) ` `            ``return` `1; ` ` `  `        ``// If there is one row or one ` `        ``// column then there is only  ` `        ``// one way of filling the matrix ` `        ``else` `if` `((n + m) % 2 == 1 && k == -1) ` `            ``return` `0; ` ` `  `        ``// If the above cases are not  ` `        ``// followed then we find ways ` `        ``// to fill the n - 1 rows and ` `        ``// m - 1 columns which is ` `        ``// 2 ^ ((m-1)*(n-1)). ` `        ``return` `(modPower(modPower((``long``)2, n - 1,  ` `                         ``mod), m - 1, mod) % mod); ` `                                     `  `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 2, m = 7, k = 1; ` `        ``Console.WriteLine(countWays(n, m, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `>= 1; ` `    ``} ` `    ``return` `\$ret``; ` `} ` ` `  `// Function calculating the answer ` `function` `countWays(``\$n``, ``\$m``, ``\$k``) ` `{ ` `    ``global` `\$mod``; ` `     `  `    ``// if sum of numbers of rows ` `    ``// and columns is odd i.e  ` `    ``// (n + m) % 2 == 1 and k = -1  ` `    ``// then there are 0 ways of  ` `    ``// filiing the matrix. ` `    ``if` `(``\$k` `== -1 ``and` `(``\$n` `+ ``\$m``) % 2 == 1) ` `        ``return` `0; ` ` `  `    ``// If there is one row or ` `    ``// one column then there ` `    ``// is only one way of  ` `    ``// filling the matrix ` `    ``if` `(``\$n` `== 1 ``or` `\$m` `== 1) ` `        ``return` `1; ` ` `  `    ``// If the above cases are ` `    ``// not followed then we ` `    ``// find ways to fill the  ` `    ``// n - 1 rows and m - 1 ` `    ``// columns which is  ` `    ``// 2 ^ ((m-1)*(n-1)). ` `    ``return` `(modPower(modPower(2, ``\$n` `- 1), ` `                         ``\$m` `- 1) % ``\$mod``); ` `} ` ` `  `    ``// Driver Code ` `    ``\$n` `= 2;  ` `    ``\$m` `= 7;  ` `    ``\$k` `= 1; ` `    ``echo` `countWays(``\$n``, ``\$m``, ``\$k``); ` `     `  `// This code is contributed by anuj_67. ` `?> `

Output:

```64
```

The time complexity of above solution is .

## tags:

Combinatorial Matrix Matrix Combinatorial