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# Program to find number of elements in matrix follows row column criteria in Python

Suppose we have a binary matrix; we have to find the number of elements in matrix that follows the following rules −

matrix[r, c] = 1

matrix[r, j] = 0 for every j when j is not equal to c and matrix[i, c] = 0 for every i when i is not equal to r.

So, if the input is like

0 | 0 | 1 |

1 | 0 | 0 |

0 | 1 | 0 |

then the output will be 3, because we have cells (0,2), (1,0) and (2,1) those meet the criteria.

To solve this, we will follow these steps −

if matrix is empty, then

return 0

row := a list of sum of all row entries in matrix

col := a list of sum of all column entries in matrix

m := row count of matrix

n := column count of matrix

res := 0

for r in range 0 to m - 1, do

for c in range 0 to n - 1, do

if matrix[r, c] is 1 and row[r] is 1 and col[c] is also 1, then

res := res + 1

return res

## Example

Let us see the following implementation to get better understanding

def solve(matrix): if not matrix: return 0 row = [sum(r) for r in matrix] col = [sum(c) for c in zip(*matrix)] m, n = len(matrix), len(matrix[0]) res = 0 for r in range(m): for c in range(n): if matrix[r][c] == 1 and row[r] == 1 and col[c] == 1: res += 1 return res matrix = [ [0, 0, 1], [1, 0, 0], [0, 1, 0] ] print(solve(matrix))

## Input

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

## Output

3

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