Program to find number of special positions in a binary matrix using Python

Suppose we have a binary matrix of order m x n, we have to find the number of special positions in the matrix. A position (i,j) is a special position when mat[i,j] = 1 and all other elements in row i and column j are 0.

So, if the input is like

then the output will be 3, here the special positions are (0, 0), (1,2) and (3,1).

To solve this, we will follow these steps −

• special := 0

• for i in range 0 to row count of matrix, do

  • if number of 1s in row matrix[i] is 1, then

    • numOfOne := 0

    • indexOfOne := position of 1 in matrix[i]

    • for j in range 0 to column size of matrix, do

      • if matrix[j, indexOfOne] is same as 1, then

        • numOfOne := numOfOne + 1

      • if numOfOne > 1, then

        • come out from the loop

    • if numOfOne is same as 1, then

      • special := special + 1

• return special

Example (Python)

Let us see the following implementation to get better understanding −

 Live Demo

def solve(matrix):
   special = 0
   for i in range(len(matrix)):
      if matrix[i].count(1) == 1:
         numOfOne = 0
         indexOfOne = matrix[i].index(1)
         for j in range(len(matrix)):
            if matrix[j][indexOfOne] == 1:
               numOfOne += 1
            if numOfOne > 1:
               break

         if numOfOne == 1:
            special += 1

   return special

matrix = [[1,0,0,0,0],
   [0,0,1,0,0],
   [0,0,0,1,1],
   [0,1,0,0,0]]
print(solve(matrix))

Input

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

Output

3