1252\. Cells with Odd Values in a Matrix

Easy

There is an `m x n` matrix that is initialized to all `0`'s. There is also a 2D array `indices` where each <code>indices[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> 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 <code>r<sub>i</sub></code>.
2.  Increment **all** the cells on column <code>c<sub>i</sub></code>.

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:**

![](https://assets.leetcode.com/uploads/2019/10/30/e1.png)

**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:**

![](https://assets.leetcode.com/uploads/2019/10/30/e2.png)

**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`
*   <code>0 <= r<sub>i</sub> < m</code>
*   <code>0 <= c<sub>i</sub> < n</code>

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