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2022-03-27

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cells-with-odd-values-in-a-matrix.json

{
    "data": {
        "question": {
            "questionId": "1378",
            "questionFrontendId": "1252",
            "categoryTitle": "Algorithms",
            "boundTopicId": 41394,
            "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>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>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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                    "code": "class Solution {\npublic:\n    int oddCells(int m, int n, vector<vector<int>>& indices) {\n\n    }\n};",
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                    "code": "class Solution {\n    public int oddCells(int m, int n, int[][] indices) {\n\n    }\n}",
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                    "code": "class Solution(object):\n    def oddCells(self, m, n, indices):\n        \"\"\"\n        :type m: int\n        :type n: int\n        :type indices: List[List[int]]\n        :rtype: int\n        \"\"\"",
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                    "code": "public class Solution {\n    public int OddCells(int m, int n, int[][] indices) {\n\n    }\n}",
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                    "code": "class Solution {\n    func oddCells(_ m: Int, _ n: Int, _ indices: [[Int]]) -> Int {\n\n    }\n}",
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                    "code": "impl Solution {\n    pub fn odd_cells(m: i32, n: i32, indices: Vec<Vec<i32>>) -> i32 {\n\n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $m\n     * @param Integer $n\n     * @param Integer[][] $indices\n     * @return Integer\n     */\n    function oddCells($m, $n, $indices) {\n\n    }\n}",
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            "hints": [
                "Simulation : With small constraints, it is possible to apply changes to each row and column and count odd cells after applying it.",
                "You can accumulate the number you should add to each row and column and then you can count the number of odd cells."
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