leetcode-problemset/leetcode-cn/originData/find-x-sum-of-all-k-long-subarrays-i.json

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            "title": "Find X-Sum of All K-Long Subarrays I",
            "titleSlug": "find-x-sum-of-all-k-long-subarrays-i",
            "content": "<p>You are given an array <code>nums</code> of <code>n</code> integers and two integers <code>k</code> and <code>x</code>.</p>\n\n<p>The <strong>x-sum</strong> of an array is calculated by the following procedure:</p>\n\n<ul>\n\t<li>Count the occurrences of all elements in the array.</li>\n\t<li>Keep only the occurrences of the top <code>x</code> most frequent elements. If two elements have the same number of occurrences, the element with the <strong>bigger</strong> value is considered more frequent.</li>\n\t<li>Calculate the sum of the resulting array.</li>\n</ul>\n\n<p><strong>Note</strong> that if an array has less than <code>x</code> distinct elements, its <strong>x-sum</strong> is the sum of the array.</p>\n\n<p>Return an integer array <code>answer</code> of length <code>n - k + 1</code> where <code>answer[i]</code> is the <strong>x-sum</strong> of the <span data-keyword=\"subarray-nonempty\">subarray</span> <code>nums[i..i + k - 1]</code>.</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 = [1,1,2,2,3,4,2,3], k = 6, x = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[6,10,12]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>For subarray <code>[1, 1, 2, 2, 3, 4]</code>, only elements 1 and 2 will be kept in the resulting array. Hence, <code>answer[0] = 1 + 1 + 2 + 2</code>.</li>\n\t<li>For subarray <code>[1, 2, 2, 3, 4, 2]</code>, only elements 2 and 4 will be kept in the resulting array. Hence, <code>answer[1] = 2 + 2 + 2 + 4</code>. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times.</li>\n\t<li>For subarray <code>[2, 2, 3, 4, 2, 3]</code>, only elements 2 and 3 are kept in the resulting array. Hence, <code>answer[2] = 2 + 2 + 2 + 3 + 3</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 = [3,8,7,8,7,5], k = 2, x = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[11,15,15,15,12]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>Since <code>k == x</code>, <code>answer[i]</code> is equal to the sum of the subarray <code>nums[i..i + k - 1]</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n\t<li><code>1 &lt;= x &lt;= k &lt;= nums.length</code></li>\n</ul>\n",
            "translatedTitle": "计算子数组的 x-sum I",
            "translatedContent": "<p>给你一个由 <code>n</code> 个整数组成的数组 <code>nums</code>，以及两个整数 <code>k</code> 和 <code>x</code>。</p>\n\n<p>数组的 <strong>x-sum</strong> 计算按照以下步骤进行：</p>\n\n<ul>\n\t<li>统计数组中所有元素的出现次数。</li>\n\t<li>仅保留出现次数最多的前 <code>x</code> 个元素的每次出现。如果两个元素的出现次数相同，则数值<strong> 较大 </strong>的元素被认为出现次数更多。</li>\n\t<li>计算结果数组的和。</li>\n</ul>\n\n<p><strong>注意</strong>，如果数组中的不同元素少于 <code>x</code> 个，则其 <strong>x-sum</strong> 是数组的元素总和。</p>\n\n<p>返回一个长度为 <code>n - k + 1</code> 的整数数组 <code>answer</code>，其中 <code>answer[i]</code> 是 <span data-keyword=\"subarray-nonempty\">子数组</span> <code>nums[i..i + k - 1]</code> 的 <strong>x-sum</strong>。</p>\n\n<p><strong>子数组</strong> 是数组内的一个连续<b> 非空</b> 的元素序列。</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><strong>输入：</strong><span class=\"example-io\">nums = [1,1,2,2,3,4,2,3], k = 6, x = 2</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">[6,10,12]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>对于子数组 <code>[1, 1, 2, 2, 3, 4]</code>，只保留元素 1 和 2。因此，<code>answer[0] = 1 + 1 + 2 + 2</code>。</li>\n\t<li>对于子数组 <code>[1, 2, 2, 3, 4, 2]</code>，只保留元素 2 和 4。因此，<code>answer[1] = 2 + 2 + 2 + 4</code>。注意 4 被保留是因为其数值大于出现其他出现次数相同的元素（3 和 1）。</li>\n\t<li>对于子数组 <code>[2, 2, 3, 4, 2, 3]</code>，只保留元素 2 和 3。因此，<code>answer[2] = 2 + 2 + 2 + 3 + 3</code>。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">nums = [3,8,7,8,7,5], k = 2, x = 2</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">[11,15,15,15,12]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>由于 <code>k == x</code>，<code>answer[i]</code> 等于子数组 <code>nums[i..i + k - 1]</code> 的总和。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n\t<li><code>1 &lt;= x &lt;= k &lt;= nums.length</code></li>\n</ul>\n",
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