mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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207 lines
25 KiB
JSON
207 lines
25 KiB
JSON
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"title": "Frequency of the Most Frequent Element",
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"content": "<p>The <strong>frequency</strong> of an element is the number of times it occurs in an array.</p>\n\n<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. In one operation, you can choose an index of <code>nums</code> and increment the element at that index by <code>1</code>.</p>\n\n<p>Return <em>the <strong>maximum possible frequency</strong> of an element after performing <strong>at most</strong> </em><code>k</code><em> operations</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,4], k = 5\n<strong>Output:</strong> 3<strong>\nExplanation:</strong> Increment the first element three times and the second element two times to make nums = [4,4,4].\n4 has a frequency of 3.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,4,8,13], k = 5\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> There are multiple optimal solutions:\n- Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.\n- Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.\n- Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,9,6], k = 2\n<strong>Output:</strong> 1\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= 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 <= k <= 10<sup>5</sup></code></li>\n</ul>\n",
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"translatedTitle": "最高频元素的频数",
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"translatedContent": "<p>元素的 <strong>频数</strong> 是该元素在一个数组中出现的次数。</p>\n\n<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> 。在一步操作中，你可以选择 <code>nums</code> 的一个下标，并将该下标对应元素的值增加 <code>1</code> 。</p>\n\n<p>执行最多 <code>k</code> 次操作后，返回数组中最高频元素的 <strong>最大可能频数</strong> <em>。</em></p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,4], k = 5\n<strong>输出：</strong>3<strong>\n解释：</strong>对第一个元素执行 3 次递增操作，对第二个元素执 2 次递增操作，此时 nums = [4,4,4] 。\n4 是数组中最高频元素，频数是 3 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,4,8,13], k = 5\n<strong>输出：</strong>2\n<strong>解释：</strong>存在多种最优解决方案：\n- 对第一个元素执行 3 次递增操作，此时 nums = [4,4,8,13] 。4 是数组中最高频元素，频数是 2 。\n- 对第二个元素执行 4 次递增操作，此时 nums = [1,8,8,13] 。8 是数组中最高频元素，频数是 2 。\n- 对第三个元素执行 5 次递增操作，此时 nums = [1,4,13,13] 。13 是数组中最高频元素，频数是 2 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,9,6], k = 2\n<strong>输出：</strong>1\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= 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 <= k <= 10<sup>5</sup></code></li>\n</ul>\n",
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