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Abhishek Chaudhary

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# Cells with Odd Values in a Matrix

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:

1. Increment all the cells on row `ri`.
2. 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?

SOLUTION:

``````class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
cols = [0] * n
rows = [0] * m
ctr = 0
for r, c in indices:
rows[r] += 1
cols[c] += 1
for i in range(n):
for j in range(m):
if (cols[i] + rows[j]) % 2 == 1:
ctr += 1
return ctr
``````