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.

In simple terms, we need to find cells that contain 1 and are the only 1 in both their row and column.

Example Input

So, if the input is like

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

Algorithm

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

Implementation

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))

The output of the above code is ?

3

How It Works

The algorithm works by first calculating the sum of each row and column. If a cell contains 1 and both its row sum and column sum equal 1, it means that cell is the only 1 in both its row and column, satisfying our criteria.

# Example with detailed steps
def solve_detailed(matrix):
    if not matrix:
        return 0
    
    # Calculate row sums
    row = [sum(r) for r in matrix]
    print("Row sums:", row)
    
    # Calculate column sums  
    col = [sum(c) for c in zip(*matrix)]
    print("Column sums:", col)
    
    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:
                print(f"Found valid cell at ({r},{c})")
                res += 1
    
    return res

matrix = [
    [0, 0, 1],
    [1, 0, 0], 
    [0, 1, 0]
]
print("Result:", solve_detailed(matrix))

Row sums: [1, 1, 1]
Column sums: [1, 1, 1]
Found valid cell at (0,2)
Found valid cell at (1,0)
Found valid cell at (2,1)
Result: 3

Conclusion

This solution efficiently finds matrix elements that are the only 1 in both their row and column by pre-computing row and column sums. The time complexity is O(m×n) where m and n are the matrix dimensions.