leetcode-problemset/leetcode-cn/originData/cells-with-odd-values-in-a-matrix.json

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            "title": "Cells with Odd Values in a Matrix",
            "titleSlug": "cells-with-odd-values-in-a-matrix",
            "content": "<p>There is an <code>m x n</code> matrix that is initialized to all <code>0</code>&#39;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>\n\n<p>For each location <code>indices[i]</code>, do <strong>both</strong> of the following:</p>\n\n<ol>\n\t<li>Increment <strong>all</strong> the cells on row <code>r<sub>i</sub></code>.</li>\n\t<li>Increment <strong>all</strong> the cells on column <code>c<sub>i</sub></code>.</li>\n</ol>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/10/30/e1.png\" style=\"width: 600px; height: 118px;\" />\n<pre>\n<strong>Input:</strong> m = 2, n = 3, indices = [[0,1],[1,1]]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> Initial matrix = [[0,0,0],[0,0,0]].\nAfter applying first increment it becomes [[1,2,1],[0,1,0]].\nThe final matrix is [[1,3,1],[1,3,1]], which contains 6 odd numbers.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/10/30/e2.png\" style=\"width: 600px; height: 150px;\" />\n<pre>\n<strong>Input:</strong> m = 2, n = 2, indices = [[1,1],[0,0]]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Final matrix = [[2,2],[2,2]]. There are no odd numbers in the final matrix.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= m, n &lt;= 50</code></li>\n\t<li><code>1 &lt;= indices.length &lt;= 100</code></li>\n\t<li><code>0 &lt;= r<sub>i</sub> &lt; m</code></li>\n\t<li><code>0 &lt;= c<sub>i</sub> &lt; n</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<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>\n",
            "translatedTitle": "奇数值单元格的数目",
            "translatedContent": "<p>给你一个 <code>m x n</code> 的矩阵，最开始的时候，每个单元格中的值都是 <code>0</code>。</p>\n\n<p>另有一个二维索引数组 <code>indices</code>，<code>indices[i] = [ri, ci]</code> 指向矩阵中的某个位置，其中 <code>ri</code> 和 <code>ci</code> 分别表示指定的行和列（<strong>从 <code>0</code> 开始编号</strong>）。</p>\n\n<p>对 <code>indices[i]</code> 所指向的每个位置，应同时执行下述增量操作：</p>\n\n<ol>\n\t<li><code>r<sub>i</sub></code> 行上的所有单元格，加 <code>1</code> 。</li>\n\t<li><code>c<sub>i</sub></code> 列上的所有单元格，加 <code>1</code> 。</li>\n</ol>\n\n<p>给你 <code>m</code>、<code>n</code> 和 <code>indices</code> 。请你在执行完所有 <code>indices</code> 指定的增量操作后，返回矩阵中 <strong>奇数值单元格</strong> 的数目。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/11/06/e1.png\" style=\"height: 118px; width: 600px;\" /></p>\n\n<pre>\n<strong>输入：</strong>m = 2, n = 3, indices = [[0,1],[1,1]]\n<strong>输出：</strong>6\n<strong>解释：</strong>最开始的矩阵是 [[0,0,0],[0,0,0]]。\n第一次增量操作后得到 [[1,2,1],[0,1,0]]。\n最后的矩阵是 [[1,3,1],[1,3,1]]，里面有 6 个奇数。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/11/06/e2.png\" style=\"height: 150px; width: 600px;\" /></p>\n\n<pre>\n<strong>输入：</strong>m = 2, n = 2, indices = [[1,1],[0,0]]\n<strong>输出：</strong>0\n<strong>解释：</strong>最后的矩阵是 [[2,2],[2,2]]，里面没有奇数。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= m, n <= 50</code></li>\n\t<li><code>1 <= indices.length <= 100</code></li>\n\t<li><code>0 <= r<sub>i</sub> < m</code></li>\n\t<li><code>0 <= c<sub>i</sub> < n</code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶：</strong>你可以设计一个时间复杂度为 <code>O(n + m + indices.length)</code> 且仅用 <code>O(n + m)</code> 额外空间的算法来解决此问题吗？</p>\n",
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