|
| 1 | +<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> |
| 2 | + |
| 3 | +<p>For each location <code>indices[i]</code>, do <strong>both</strong> of the following:</p> |
| 4 | + |
| 5 | +<ol> |
| 6 | + <li>Increment <strong>all</strong> the cells on row <code>r<sub>i</sub></code>.</li> |
| 7 | + <li>Increment <strong>all</strong> the cells on column <code>c<sub>i</sub></code>.</li> |
| 8 | +</ol> |
| 9 | + |
| 10 | +<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> |
| 11 | + |
| 12 | +<p> </p> |
| 13 | +<p><strong class="example">Example 1:</strong></p> |
| 14 | +<img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e1.png" style="width: 600px; height: 118px;" /> |
| 15 | +<pre> |
| 16 | +<strong>Input:</strong> m = 2, n = 3, indices = [[0,1],[1,1]] |
| 17 | +<strong>Output:</strong> 6 |
| 18 | +<strong>Explanation:</strong> Initial matrix = [[0,0,0],[0,0,0]]. |
| 19 | +After applying first increment it becomes [[1,2,1],[0,1,0]]. |
| 20 | +The final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers. |
| 21 | +</pre> |
| 22 | + |
| 23 | +<p><strong class="example">Example 2:</strong></p> |
| 24 | +<img alt="" src="https://assets.leetcode.com/uploads/2019/10/30/e2.png" style="width: 600px; height: 150px;" /> |
| 25 | +<pre> |
| 26 | +<strong>Input:</strong> m = 2, n = 2, indices = [[1,1],[0,0]] |
| 27 | +<strong>Output:</strong> 0 |
| 28 | +<strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix. |
| 29 | +</pre> |
| 30 | + |
| 31 | +<p> </p> |
| 32 | +<p><strong>Constraints:</strong></p> |
| 33 | + |
| 34 | +<ul> |
| 35 | + <li><code>1 <= m, n <= 50</code></li> |
| 36 | + <li><code>1 <= indices.length <= 100</code></li> |
| 37 | + <li><code>0 <= r<sub>i</sub> < m</code></li> |
| 38 | + <li><code>0 <= c<sub>i</sub> < n</code></li> |
| 39 | +</ul> |
| 40 | + |
| 41 | +<p> </p> |
| 42 | +<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> |
0 commit comments