Count zeros in a row wise and column wise sorted matrix - GeeksforGeeks

Count zeros in a row wise and column wise sorted matrix - GeeksforGeeks
Given a N x N binary matrix (elements in matrix can be either 1 or 0) where each row and column of the matrix is sorted in ascending order, count number of 0s present in it.
Expected time complexity is O(N).

The idea is very simple. We start from the bottom-left corner of the matrix and repeat below steps until we find the top or right edge of the matrix.

1. Decrement row index until we find a 0.
2. Add number of 0s in current column i.e. current row index + 1 to the result and move right to next column (Increment col index by 1).

The above logic will work since the matrix is row-wise and column-wise sorted. The logic will also work for any matrix containing non-negative integers.

int countZeroes(int mat[N][N])

{

    // start from bottom-left corner of the matrix

    int row = N - 1, col = 0;

 

    // stores number of zeroes in the matrix

    int count = 0;

 

    while (col < N)

    {

        // move up until you find a 0

        while (mat[row][col])

 

            // if zero is not found in current column,

            // we are done

            if (--row < 0)

                return count;

 

        // add 0s present in current column to result

        count += (row + 1);

 

        // move right to next column

        col++;

    }

 

    return count;

}

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