<p>There is an <code>m x n</code> matrix that is initialized to all <code>0</code>'s. There is also a 2D array <code>indices</code> where each <code>indices[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> represents a <strong>0-indexed location</strong> to perform some increment operations on the matrix.</p> <p>For each location <code>indices[i]</code>, do <strong>both</strong> of the following:</p> <ol> <li>Increment <strong>all</strong> the cells on row <code>r<sub>i</sub></code>.</li> <li>Increment <strong>all</strong> the cells on column <code>c<sub>i</sub></code>.</li> </ol> <p>Given <code>m</code>, <code>n</code>, and <code>indices</code>, return <em>the <strong>number of odd-valued cells</strong> in the matrix after applying the increment to all locations in </em><code>indices</code>.</p> <p> </p> <p><strong>Example 1:</strong></p> <img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e1.png" style="width: 600px; height: 118px;" /> <pre> <strong>Input:</strong> m = 2, n = 3, indices = [[0,1],[1,1]] <strong>Output:</strong> 6 <strong>Explanation:</strong> 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. </pre> <p><strong>Example 2:</strong></p> <img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e2.png" style="width: 600px; height: 150px;" /> <pre> <strong>Input:</strong> m = 2, n = 2, indices = [[1,1],[0,0]] <strong>Output:</strong> 0 <strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix. </pre> <p> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 <= m, n <= 50</code></li> <li><code>1 <= indices.length <= 100</code></li> <li><code>0 <= r<sub>i</sub> < m</code></li> <li><code>0 <= c<sub>i</sub> < n</code></li> </ul> <p> </p> <p><strong>Follow up:</strong> Could you solve this in <code>O(n + m + indices.length)</code> time with only <code>O(n + m)</code> extra space?</p>