leetcode-problemset/leetcode-cn/originData/maximize-happiness-of-selected-children.json

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        "question": {
            "questionId": "3351",
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            "categoryTitle": "Algorithms",
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            "title": "Maximize Happiness of Selected Children",
            "titleSlug": "maximize-happiness-of-selected-children",
            "content": "<p>You are given an array <code>happiness</code> of length <code>n</code>, and a <strong>positive</strong> integer <code>k</code>.</p>\n\n<p>There are <code>n</code> children standing in a queue, where the <code>i<sup>th</sup></code> child has <strong>happiness value</strong> <code>happiness[i]</code>. You want to select <code>k</code> children from these <code>n</code> children in <code>k</code> turns.</p>\n\n<p>In each turn, when you select a child, the <strong>happiness value</strong> of all the children that have <strong>not</strong> been selected till now decreases by <code>1</code>. Note that the happiness value <strong>cannot</strong> become negative and gets decremented <strong>only</strong> if it is positive.</p>\n\n<p>Return <em>the <strong>maximum</strong> sum of the happiness values of the selected children you can achieve by selecting </em><code>k</code> <em>children</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> happiness = [1,2,3], k = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> We can pick 2 children in the following way:\n- Pick the child with the happiness value == 3. The happiness value of the remaining children becomes [0,1].\n- Pick the child with the happiness value == 1. The happiness value of the remaining child becomes [0]. Note that the happiness value cannot become less than 0.\nThe sum of the happiness values of the selected children is 3 + 1 = 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> happiness = [1,1,1,1], k = 2\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> We can pick 2 children in the following way:\n- Pick any child with the happiness value == 1. The happiness value of the remaining children becomes [0,0,0].\n- Pick the child with the happiness value == 0. The happiness value of the remaining child becomes [0,0].\nThe sum of the happiness values of the selected children is 1 + 0 = 1.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> happiness = [2,3,4,5], k = 1\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> We can pick 1 child in the following way:\n- Pick the child with the happiness value == 5. The happiness value of the remaining children becomes [1,2,3].\nThe sum of the happiness values of the selected children is 5.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == happiness.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= happiness[i] &lt;= 10<sup>8</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n</code></li>\n</ul>\n",
            "translatedTitle": "幸福值最大化的选择方案",
            "translatedContent": "<p>给你一个长度为 <code>n</code> 的数组 <code>happiness</code> ，以及一个<strong> 正整数 </strong><code>k</code> 。</p>\n\n<p><code>n</code> 个孩子站成一队，其中第 <code>i</code> 个孩子的 <strong>幸福值</strong> 是<strong> </strong><code>happiness[i]</code> 。你计划组织 <code>k</code> 轮筛选从这 <code>n</code> 个孩子中选出 <code>k</code> 个孩子。</p>\n\n<p>在每一轮选择一个孩子时，所有<strong> 尚未 </strong>被选中的孩子的 <strong>幸福值 </strong>将减少 <code>1</code> 。注意，幸福值<strong> 不能 </strong>变成负数，且只有在它是正数的情况下才会减少。</p>\n\n<p>选择 <code>k</code> 个孩子，并使你选中的孩子幸福值之和最大，返回你能够得到的<strong> </strong><strong>最大值</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>happiness = [1,2,3], k = 2\n<strong>输出：</strong>4\n<strong>解释：</strong>按以下方式选择 2 个孩子：\n- 选择幸福值为 3 的孩子。剩余孩子的幸福值变为 [0,1] 。\n- 选择幸福值为 1 的孩子。剩余孩子的幸福值变为 [0] 。注意幸福值不能小于 0 。\n所选孩子的幸福值之和为 3 + 1 = 4 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>happiness = [1,1,1,1], k = 2\n<strong>输出：</strong>1\n<strong>解释：</strong>按以下方式选择 2 个孩子：\n- 选择幸福值为 1 的任意一个孩子。剩余孩子的幸福值变为 [0,0,0] 。\n- 选择幸福值为 0 的孩子。剩余孩子的幸福值变为 [0,0] 。\n所选孩子的幸福值之和为 1 + 0 = 1 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>happiness = [2,3,4,5], k = 1\n<strong>输出：</strong>5\n<strong>解释：</strong>按以下方式选择 1 个孩子：\n- 选择幸福值为 5 的孩子。剩余孩子的幸福值变为 [1,2,3] 。\n所选孩子的幸福值之和为 5 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == happiness.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= happiness[i] &lt;= 10<sup>8</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= n</code></li>\n</ul>\n",
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                "Since all the unselected numbers are decreasing at the same rate, we should greedily select <code>k</code> largest values.",
                "The <code>i<sup>th</code> largest number (<code>i = 1, 2, 3,…k</code>) should decrease by <code>(i - 1)</code> when it is picked.",
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