Special Positions in a Binary Matrix
To solve this problem, you need to count how many times each row and each column in the matrix has a 1, then iterate over the matrix again and for every cell that is 1, you check if its row and column counts are both 1. If they are, increment a special positions counter.
Here is the Python solution:


This solution works in O(m*n) time and O(m + n) space, where m is the number of rows in the matrix and n is the number of columns.
Q&A
Why expected 3 in [[0,0,0,0,0],[1,0,0,0,0],[0,1,0,0,0],[0,0,1,0,0],[0,0,0,1,1]], if right answer is 4?
The matrix you’ve provided is:
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 1
In this case, the special positions (where the cell value is 1 and all other elements in its row and column are 0) are:
 (1, 0) with value 1
 (2, 1) with value 1
 (3, 2) with value 1
 (4, 3) with value 1
The position (4, 4) is not special because even though it has value 1, there is another 1 in its column at position (4, 3).
So, there are indeed 4 special positions, not 3. If the expected answer is 3, there might be a mistake in the problem statement or the test case.