leetcode-problemset/leetcode-cn/originData/total-cost-to-hire-k-workers.json

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            "title": "Total Cost to Hire K Workers",
            "titleSlug": "total-cost-to-hire-k-workers",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>costs</code> where <code>costs[i]</code> is the cost of hiring the <code>i<sup>th</sup></code> worker.</p>\n\n<p>You are also given two integers <code>k</code> and <code>candidates</code>. We want to hire exactly <code>k</code> workers according to the following rules:</p>\n\n<ul>\n\t<li>You will run <code>k</code> sessions and hire exactly one worker in each session.</li>\n\t<li>In each hiring session, choose the worker with the lowest cost from either the first <code>candidates</code> workers or the last <code>candidates</code> workers. Break the tie by the smallest index.\n\t<ul>\n\t\t<li>For example, if <code>costs = [3,2,7,7,1,2]</code> and <code>candidates = 2</code>, then in the first hiring session, we will choose the <code>4<sup>th</sup></code> worker because they have the lowest cost <code>[<u>3,2</u>,7,7,<u><strong>1</strong>,2</u>]</code>.</li>\n\t\t<li>In the second hiring session, we will choose <code>1<sup>st</sup></code> worker because they have the same lowest cost as <code>4<sup>th</sup></code> worker but they have the smallest index <code>[<u>3,<strong>2</strong></u>,7,<u>7,2</u>]</code>. Please note that the indexing may be changed in the process.</li>\n\t</ul>\n\t</li>\n\t<li>If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.</li>\n\t<li>A worker can only be chosen once.</li>\n</ul>\n\n<p>Return <em>the total cost to hire exactly </em><code>k</code><em> workers.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4\n<strong>Output:</strong> 11\n<strong>Explanation:</strong> We hire 3 workers in total. The total cost is initially 0.\n- In the first hiring round we choose the worker from [<u>17,12,10,2</u>,7,<u>2,11,20,8</u>]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.\n- In the second hiring round we choose the worker from [<u>17,12,10,7</u>,<u>2,11,20,8</u>]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.\n- In the third hiring round we choose the worker from [<u>17,12,10,7,11,20,8</u>]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.\nThe total hiring cost is 11.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> costs = [1,2,4,1], k = 3, candidates = 3\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> We hire 3 workers in total. The total cost is initially 0.\n- In the first hiring round we choose the worker from [<u>1,2,4,1</u>]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.\n- In the second hiring round we choose the worker from [<u>2,4,1</u>]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.\n- In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [<u>2,4</u>]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.\nThe total hiring cost is 4.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= costs.length &lt;= 10<sup>5 </sup></code></li>\n\t<li><code>1 &lt;= costs[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k, candidates &lt;= costs.length</code></li>\n</ul>\n",
            "translatedTitle": "雇佣 K 位工人的总代价",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>costs</code>&nbsp;，其中&nbsp;<code>costs[i]</code>&nbsp;是雇佣第 <code>i</code>&nbsp;位工人的代价。</p>\n\n<p>同时给你两个整数&nbsp;<code>k</code> 和&nbsp;<code>candidates</code>&nbsp;。我们想根据以下规则恰好雇佣&nbsp;<code>k</code>&nbsp;位工人：</p>\n\n<ul>\n\t<li>总共进行&nbsp;<code>k</code>&nbsp;轮雇佣，且每一轮恰好雇佣一位工人。</li>\n\t<li>在每一轮雇佣中，从最前面 <code>candidates</code>&nbsp;和最后面 <code>candidates</code>&nbsp;人中选出代价最小的一位工人，如果有多位代价相同且最小的工人，选择下标更小的一位工人。\n\t<ul>\n\t\t<li>比方说，<code>costs = [3,2,7,7,1,2]</code> 且&nbsp;<code>candidates = 2</code>&nbsp;，第一轮雇佣中，我们选择第&nbsp;<code>4</code>&nbsp;位工人，因为他的代价最小&nbsp;<code>[<em>3,2</em>,7,7,<em><strong>1</strong>,2</em>]</code>&nbsp;。</li>\n\t\t<li>第二轮雇佣，我们选择第&nbsp;<code>1</code>&nbsp;位工人，因为他们的代价与第&nbsp;<code>4</code>&nbsp;位工人一样都是最小代价，而且下标更小，<code>[<em>3,<strong>2</strong></em>,7,<em>7,2</em>]</code>&nbsp;。注意每一轮雇佣后，剩余工人的下标可能会发生变化。</li>\n\t</ul>\n\t</li>\n\t<li>如果剩余员工数目不足 <code>candidates</code>&nbsp;人，那么下一轮雇佣他们中代价最小的一人，如果有多位代价相同且最小的工人，选择下标更小的一位工人。</li>\n\t<li>一位工人只能被选择一次。</li>\n</ul>\n\n<p>返回雇佣恰好<em>&nbsp;</em><code>k</code>&nbsp;位工人的总代价。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4\n<b>输出：</b>11\n<b>解释：</b>我们总共雇佣 3 位工人。总代价一开始为 0 。\n- 第一轮雇佣，我们从 [<strong><em>17,12,10,2</em></strong>,7,<strong><em>2,11,20,8</em></strong>] 中选择。最小代价是 2 ，有两位工人，我们选择下标更小的一位工人，即第 3 位工人。总代价是 0 + 2 = 2 。\n- 第二轮雇佣，我们从 [<strong><em>17,12,10,7</em></strong>,<strong><em>2,11,20,8</em></strong>] 中选择。最小代价是 2 ，下标为 4 ，总代价是 2 + 2 = 4 。\n- 第三轮雇佣，我们从 [<strong><em>17,12,10,7,11,20,8</em></strong>] 中选择，最小代价是 7 ，下标为 3 ，总代价是 4 + 7 = 11 。注意下标为 3 的工人同时在最前面和最后面 4 位工人中。\n总雇佣代价是 11 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>costs = [1,2,4,1], k = 3, candidates = 3\n<b>输出：</b>4\n<b>解释：</b>我们总共雇佣 3 位工人。总代价一开始为 0 。\n- 第一轮雇佣，我们从 [<strong><em>1,2,4,1</em></strong>] 中选择。最小代价为 1 ，有两位工人，我们选择下标更小的一位工人，即第 0 位工人，总代价是 0 + 1 = 1 。注意，下标为 1 和 2 的工人同时在最前面和最后面 3 位工人中。\n- 第二轮雇佣，我们从 [<strong><em>2,4,1</em></strong>] 中选择。最小代价为 1 ，下标为 2 ，总代价是 1 + 1 = 2 。\n- 第三轮雇佣，少于 3 位工人，我们从剩余工人 [<strong><em>2,4</em></strong>] 中选择。最小代价是 2 ，下标为 0 。总代价为 2 + 2 = 4 。\n总雇佣代价是 4 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= costs.length &lt;= 10<sup>5 </sup></code></li>\n\t<li><code>1 &lt;= costs[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k, candidates &lt;= costs.length</code></li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap 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