Solution.py

"""

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix.
For each location indices[i], do both of the following:

Increment all the cells on row ri.
Increment all the cells on column ci.

Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.
 
Example 1:


Input: m = 2, n = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.

Example 2:


Input: m = 2, n = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.

 
Constraints:

1 <= m, n <= 50
1 <= indices.length <= 100
0 <= ri < m
0 <= ci < n

 
Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

"""


class Solution:
    def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
        matrix = [[0 for _ in range(n)] for _ in range(m)]
        for ri, ci in indices:
            for idx in range(n):
                matrix[ri][idx] += 1
            for idx in range(m):
                matrix[idx][ci] += 1
        ret = 0
        for row in matrix:
            for elem in row:
                if elem & 1 != 0:
                    ret += 1
        return ret