leetcode-problemset/leetcode-cn/originData/the-number-of-the-smallest-unoccupied-chair.json

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            "title": "The Number of the Smallest Unoccupied Chair",
            "titleSlug": "the-number-of-the-smallest-unoccupied-chair",
            "content": "<p>There is a party where <code>n</code> friends numbered from <code>0</code> to <code>n - 1</code> are attending. There is an <strong>infinite</strong> number of chairs in this party that are numbered from <code>0</code> to <code>infinity</code>. When a friend arrives at the party, they sit on the unoccupied chair with the <strong>smallest number</strong>.</p>\n\n<ul>\n\t<li>For example, if chairs <code>0</code>, <code>1</code>, and <code>5</code> are occupied when a friend comes, they will sit on chair number <code>2</code>.</li>\n</ul>\n\n<p>When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair.</p>\n\n<p>You are given a <strong>0-indexed</strong> 2D integer array <code>times</code> where <code>times[i] = [arrival<sub>i</sub>, leaving<sub>i</sub>]</code>, indicating the arrival and leaving times of the <code>i<sup>th</sup></code> friend respectively, and an integer <code>targetFriend</code>. All arrival times are <strong>distinct</strong>.</p>\n\n<p>Return<em> the <strong>chair number</strong> that the friend numbered </em><code>targetFriend</code><em> will sit on</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> times = [[1,4],[2,3],[4,6]], targetFriend = 1\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> \n- Friend 0 arrives at time 1 and sits on chair 0.\n- Friend 1 arrives at time 2 and sits on chair 1.\n- Friend 1 leaves at time 3 and chair 1 becomes empty.\n- Friend 0 leaves at time 4 and chair 0 becomes empty.\n- Friend 2 arrives at time 4 and sits on chair 0.\nSince friend 1 sat on chair 1, we return 1.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> times = [[3,10],[1,5],[2,6]], targetFriend = 0\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \n- Friend 1 arrives at time 1 and sits on chair 0.\n- Friend 2 arrives at time 2 and sits on chair 1.\n- Friend 0 arrives at time 3 and sits on chair 2.\n- Friend 1 leaves at time 5 and chair 0 becomes empty.\n- Friend 2 leaves at time 6 and chair 1 becomes empty.\n- Friend 0 leaves at time 10 and chair 2 becomes empty.\nSince friend 0 sat on chair 2, we return 2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == times.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>times[i].length == 2</code></li>\n\t<li><code>1 &lt;= arrival<sub>i</sub> &lt; leaving<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= targetFriend &lt;= n - 1</code></li>\n\t<li>Each <code>arrival<sub>i</sub></code> time is <strong>distinct</strong>.</li>\n</ul>\n",
            "translatedTitle": "最小未被占据椅子的编号",
            "translatedContent": "<p>有 <code>n</code> 个朋友在举办一个派对，这些朋友从 <code>0</code> 到 <code>n - 1</code> 编号。派对里有 <strong>无数</strong> 张椅子，编号为 <code>0</code> 到 <code>infinity</code> 。当一个朋友到达派对时，他会占据 <strong>编号最小</strong> 且未被占据的椅子。</p>\n\n<ul>\n\t<li>比方说，当一个朋友到达时，如果椅子 <code>0</code> ，<code>1</code> 和 <code>5</code> 被占据了，那么他会占据 <code>2</code> 号椅子。</li>\n</ul>\n\n<p>当一个朋友离开派对时，他的椅子会立刻变成未占据状态。如果同一时刻有另一个朋友到达，可以立即占据这张椅子。</p>\n\n<p>给你一个下标从 <strong>0</strong> 开始的二维整数数组 <code>times</code> ，其中 <code>times[i] = [arrival<sub>i</sub>, leaving<sub>i</sub>]</code> 表示第 <code>i</code> 个朋友到达和离开的时刻，同时给你一个整数 <code>targetFriend</code> 。所有到达时间 <strong>互不相同</strong> 。</p>\n\n<p>请你返回编号为 <code>targetFriend</code> 的朋友占据的 <strong>椅子编号</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>times = [[1,4],[2,3],[4,6]], targetFriend = 1\n<b>输出：</b>1\n<b>解释：</b>\n- 朋友 0 时刻 1 到达，占据椅子 0 。\n- 朋友 1 时刻 2 到达，占据椅子 1 。\n- 朋友 1 时刻 3 离开，椅子 1 变成未占据。\n- 朋友 0 时刻 4 离开，椅子 0 变成未占据。\n- 朋友 2 时刻 4 到达，占据椅子 0 。\n朋友 1 占据椅子 1 ，所以返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>times = [[3,10],[1,5],[2,6]], targetFriend = 0\n<b>输出：</b>2\n<b>解释：</b>\n- 朋友 1 时刻 1 到达，占据椅子 0 。\n- 朋友 2 时刻 2 到达，占据椅子 1 。\n- 朋友 0 时刻 3 到达，占据椅子 2 。\n- 朋友 1 时刻 5 离开，椅子 0 变成未占据。\n- 朋友 2 时刻 6 离开，椅子 1 变成未占据。\n- 朋友 0 时刻 10 离开，椅子 2 变成未占据。\n朋友 0 占据椅子 2 ，所以返回 2 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == times.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>4</sup></code></li>\n\t<li><code>times[i].length == 2</code></li>\n\t<li><code>1 &lt;= arrival<sub>i</sub> &lt; leaving<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= targetFriend &lt;= n - 1</code></li>\n\t<li>每个 <code>arrival<sub>i</sub></code> 时刻 <strong>互不相同</strong> 。</li>\n</ul>\n",
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