mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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180 lines
17 KiB
JSON
180 lines
17 KiB
JSON
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"questionId": "1378",
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"title": "Cells with Odd Values in a Matrix",
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"content": "<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>\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> </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> </p>\n<p><strong>Constraints:</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<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",
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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": "# @param {Integer} m\n# @param {Integer} n\n# @param {Integer[][]} indices\n# @return {Integer}\ndef odd_cells(m, n, indices)\n \nend",
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"code": "object Solution {\n def oddCells(m: Int, n: Int, indices: Array[Array[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": "-spec odd_cells(M :: integer(), N :: integer(), Indices :: [[integer()]]) -> integer().\nodd_cells(M, N, Indices) ->\n .",
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"code": "defmodule Solution do\n @spec odd_cells(m :: integer, n :: integer, indices :: [[integer]]) :: integer\n def odd_cells(m, n, indices) do\n \n end\nend",
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"Simulation : With small constraints, it is possible to apply changes to each row and column and count odd cells after applying it.",
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"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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