## Description

https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/

Given `n`

and `m`

which are the dimensions of a matrix initialized by zeros and given an array `indices`

where `indices[i] = [ri, ci]`

. For each pair of `[ri, ci]`

you have to increment all cells in row `ri`

and column `ci`

by 1.

Return *the number of cells with odd values* in the matrix after applying the increment to all `indices`

.

**Example 1:**

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

**Example 2:**

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

**Constraints:**

`1 <= n <= 50`

`1 <= m <= 50`

`1 <= indices.length <= 100`

`0 <= indices[i][0] < n`

`0 <= indices[i][1] < m`

## Explanation

Simply implement as the problem description says.

## Python Solution

```
class Solution:
def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
count = 0
matrix = [[0 for j in range(m)] for i in range(n)]
for indice in indices:
r = indice[0]
c = indice[1]
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if i == r:
matrix[i][j] += 1
if j == c:
matrix[i][j] += 1
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] % 2 != 0:
count += 1
return count
```

- Time Complexity: O(N^2)
- Space Complexity: O(N^2)