leetcode-problemset/leetcode-cn/originData/maximum-frequency-of-an-element-after-performing-operations-ii.json

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            "questionId": "3640",
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            "categoryTitle": "Algorithms",
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            "title": "Maximum Frequency of an Element After Performing Operations II",
            "titleSlug": "maximum-frequency-of-an-element-after-performing-operations-ii",
            "content": "<p>You are given an integer array <code>nums</code> and two integers <code>k</code> and <code>numOperations</code>.</p>\n\n<p>You must perform an <strong>operation</strong> <code>numOperations</code> times on <code>nums</code>, where in each operation you:</p>\n\n<ul>\n\t<li>Select an index <code>i</code> that was <strong>not</strong> selected in any previous operations.</li>\n\t<li>Add an integer in the range <code>[-k, k]</code> to <code>nums[i]</code>.</li>\n</ul>\n\n<p>Return the <strong>maximum</strong> possible <span data-keyword=\"frequency-array\">frequency</span> of any element in <code>nums</code> after performing the <strong>operations</strong>.</p>\n\n<p>&nbsp;</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 = [1,4,5], k = 1, numOperations = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can achieve a maximum frequency of two by:</p>\n\n<ul>\n\t<li>Adding 0 to <code>nums[1]</code>, after which <code>nums</code> becomes <code>[1, 4, 5]</code>.</li>\n\t<li>Adding -1 to <code>nums[2]</code>, after which <code>nums</code> becomes <code>[1, 4, 4]</code>.</li>\n</ul>\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 = [5,11,20,20], k = 5, numOperations = 1</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>We can achieve a maximum frequency of two by:</p>\n\n<ul>\n\t<li>Adding 0 to <code>nums[1]</code>.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= numOperations &lt;= nums.length</code></li>\n</ul>\n",
            "translatedTitle": "执行操作后元素的最高频率 II",
            "translatedContent": "<p>给你一个整数数组&nbsp;<code>nums</code>&nbsp;和两个整数&nbsp;<code>k</code> 和&nbsp;<code>numOperations</code>&nbsp;。</p>\n\n<p>你必须对 <code>nums</code>&nbsp;执行 <strong>操作</strong>&nbsp; <code>numOperations</code>&nbsp;次。每次操作中，你可以：</p>\n\n<ul>\n\t<li>选择一个下标&nbsp;<code>i</code>&nbsp;，它在之前的操作中 <strong>没有</strong>&nbsp;被选择过。</li>\n\t<li>将 <code>nums[i]</code>&nbsp;增加范围&nbsp;<code>[-k, k]</code>&nbsp;中的一个整数。</li>\n</ul>\n\n<p>在执行完所有操作以后，请你返回 <code>nums</code>&nbsp;中出现 <strong>频率最高</strong>&nbsp;元素的出现次数。</p>\n\n<p>一个元素 <code>x</code>&nbsp;的 <strong>频率</strong>&nbsp;指的是它在数组中出现的次数。</p>\n\n<p>&nbsp;</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 = [1,4,5], k = 1, numOperations = 2</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>通过以下操作得到最高频率 2 ：</p>\n\n<ul>\n\t<li>将&nbsp;<code>nums[1]</code>&nbsp;增加 0 ，<code>nums</code> 变为&nbsp;<code>[1, 4, 5]</code>&nbsp;。</li>\n\t<li>将&nbsp;<code>nums[2]</code>&nbsp;增加 -1 ，<code>nums</code> 变为&nbsp;<code>[1, 4, 4]</code>&nbsp;。</li>\n</ul>\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 = [5,11,20,20], k = 5, numOperations = 1</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>通过以下操作得到最高频率 2 ：</p>\n\n<ul>\n\t<li>将&nbsp;<code>nums[1]</code>&nbsp;增加 0 。</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= numOperations &lt;= nums.length</code></li>\n</ul>\n",
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