mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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185 lines
26 KiB
JSON
185 lines
26 KiB
JSON
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"questionId": "3438",
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"categoryTitle": "Algorithms",
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"title": "Peaks in Array",
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"content": "<p>A <strong>peak</strong> in an array <code>arr</code> is an element that is <strong>greater</strong> than its previous and next element in <code>arr</code>.</p>\n\n<p>You are given an integer array <code>nums</code> and a 2D integer array <code>queries</code>.</p>\n\n<p>You have to process queries of two types:</p>\n\n<ul>\n\t<li><code>queries[i] = [1, l<sub>i</sub>, r<sub>i</sub>]</code>, determine the count of <strong>peak</strong> elements in the <span data-keyword=\"subarray\">subarray</span> <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code>.<!-- notionvc: 73b20b7c-e1ab-4dac-86d0-13761094a9ae --></li>\n\t<li><code>queries[i] = [2, index<sub>i</sub>, val<sub>i</sub>]</code>, change <code>nums[index<sub>i</sub>]</code> to <code><font face=\"monospace\">val<sub>i</sub></font></code>.</li>\n</ul>\n\n<p>Return an array <code>answer</code> containing the results of the queries of the first type in order.<!-- notionvc: a9ccef22-4061-4b5a-b4cc-a2b2a0e12f30 --></p>\n\n<p><strong>Notes:</strong></p>\n\n<ul>\n\t<li>The <strong>first</strong> and the <strong>last</strong> element of an array or a subarray<!-- notionvc: fcffef72-deb5-47cb-8719-3a3790102f73 --> <strong>cannot</strong> be a peak.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[0]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>First query: We change <code>nums[3]</code> to 4 and <code>nums</code> becomes <code>[3,1,4,4,5]</code>.</p>\n\n<p>Second query: The number of peaks in the <code>[3,1,4,4,5]</code> is 0.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[0,1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>First query: <code>nums[2]</code> should become 4, but it is already set to 4.</p>\n\n<p>Second query: The number of peaks in the <code>[4,1,4]</code> is 0.</p>\n\n<p>Third query: The second 4 is a peak in the <code>[4,1,4,2,1]</code>.</p>\n</div>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>3 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= queries.length <= 10<sup>5</sup></code></li>\n\t<li><code>queries[i][0] == 1</code> or <code>queries[i][0] == 2</code></li>\n\t<li>For all <code>i</code> that:\n\t<ul>\n\t\t<li><code>queries[i][0] == 1</code>: <code>0 <= queries[i][1] <= queries[i][2] <= nums.length - 1</code></li>\n\t\t<li><code>queries[i][0] == 2</code>: <code>0 <= queries[i][1] <= nums.length - 1</code>, <code>1 <= queries[i][2] <= 10<sup>5</sup></code></li>\n\t</ul>\n\t</li>\n</ul>\n",
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"translatedTitle": "数组中的峰值",
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"translatedContent": "<p>数组 <code>arr</code> 中 <strong>大于</strong> 前面和后面相邻元素的元素被称为 <strong>峰值</strong> 元素。</p>\n\n<p>给你一个整数数组 <code>nums</code> 和一个二维整数数组 <code>queries</code> 。</p>\n\n<p>你需要处理以下两种类型的操作:</p>\n\n<ul>\n\t<li><code>queries[i] = [1, l<sub>i</sub>, r<sub>i</sub>]</code> ,求出子数组 <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code> 中 <strong>峰值</strong> 元素的数目。<!-- notionvc: 73b20b7c-e1ab-4dac-86d0-13761094a9ae --></li>\n\t<li><code>queries[i] = [2, index<sub>i</sub>, val<sub>i</sub>]</code> ,将 <code>nums[index<sub>i</sub>]</code> 变为 <code><font face=\"monospace\">val<sub>i</sub></font></code><font face=\"monospace\"> 。</font></li>\n</ul>\n\n<p>请你返回一个数组 <code>answer</code> ,它依次包含每一个第一种操作的答案。<!-- notionvc: a9ccef22-4061-4b5a-b4cc-a2b2a0e12f30 --></p>\n\n<p><strong>注意:</strong></p>\n\n<ul>\n\t<li>子数组中 <strong>第一个</strong> 和 <strong>最后一个</strong> 元素都 <strong>不是</strong> 峰值元素。</li>\n</ul>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1:</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]]</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>[0]</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>第一个操作:我们将 <code>nums[3]</code> 变为 4 ,<code>nums</code> 变为 <code>[3,1,4,4,5]</code> 。</p>\n\n<p>第二个操作:<code>[3,1,4,4,5]</code> 中峰值元素的数目为 0 。</p>\n</div>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入:</b>nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]]</span></p>\n\n<p><span class=\"example-io\"><b>输出:</b>[0,1]</span></p>\n\n<p><strong>解释:</strong></p>\n\n<p>第一个操作:<code>nums[2]</code> 变为 4 ,它已经是 4 了,所以保持不变。</p>\n\n<p>第二个操作:<code>[4,1,4]</code> 中峰值元素的数目为 0 。</p>\n\n<p>第三个操作:第二个 4 是 <code>[4,1,4,2,1]</code> 中的峰值元素。</p>\n</div>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>3 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= queries.length <= 10<sup>5</sup></code></li>\n\t<li><code>queries[i][0] == 1</code> 或者 <code>queries[i][0] == 2</code></li>\n\t<li>对于所有的 <code>i</code> ,都有:\n\t<ul>\n\t\t<li><code>queries[i][0] == 1</code> :<code>0 <= queries[i][1] <= queries[i][2] <= nums.length - 1</code></li>\n\t\t<li><code>queries[i][0] == 2</code> :<code>0 <= queries[i][1] <= nums.length - 1</code>, <code>1 <= queries[i][2] <= 10<sup>5</sup></code></li>\n\t</ul>\n\t</li>\n</ul>\n",
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"Let <code>p[i]</code> be whether <code>nums[i]</code> is a peak in the original array. Namely <code>p[i] = nums[i] > nums[i - 1] && nums[i] > nums[i + 1]</code>.",
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"The answer for <code>[l<sub>i</sub>, r<sub>i</sub>]</code> is <code>p[l<sub>i</sub> + 1] + p[l<sub>i</sub> + 2] + … + p[r<sub>i</sub> - 1]</code> (note that <code>l<sub>i</sub></code> and <code>r<sub>i</sub></code> are not included).",
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