Program to check how many ways we can choose empty cells of a matrix in python

Suppose we have a N x N binary matrix where 0 is for empty cells and 1 is a blocked cells, we have to find the number of ways to choose N empty cells such that every row and every column has at least one chosen cells. If the answer is very large return result mod 10^9 + 7

So, if the input is like

then the output will be 4, as we have following configurations (where x is a selected cell) −

To solve this, we will follow these steps −

• n := size of matrix
• Define a function f() . This will take i, bs
• if i >= n, then
  • return 1
• ans := 0
• for j in range 0 to n, do
  • if matrix[i, j] is same as 0 and (2^j AND bs is same as 0) , then
    • ans := ans + f(i + 1, bs OR 2^j)
• return ans
• From the main method call and return f(0, 0)

Let us see the following implementation to get better understanding −

Example 

class Solution:
   def solve(self, matrix):
      n = len(matrix)

      def f(i, bs):
         if i >= n:
            return 1
         ans = 0
         for j in range(n):
            if matrix[i][j] == 0 and ((1 << j) & bs == 0):
               ans += f(i + 1, bs | (1 << j))
         return ans

      return f(0, 0)

ob = Solution()
matrix = [
   [0, 0, 0],
   [0, 0, 0],
   [0, 1, 0]
]
print(ob.solve(matrix))

Input

[  
[0, 0, 0],  
[0, 0, 0],  
[0, 1, 0] ]

Output

4