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feat: add weekly contest 400 (#2999)
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‎solution/1500-1599/1521.Find a Value of a Mysterious Function Closest to Target/README.md

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时间复杂度 $O(n \times \log M)$，空间复杂度 $O(\log M)$。其中 $n$ 和 $M$ 分别是数组 $arr$ 的长度和数组 $arr$ 中的最大值。
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相似题目：
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- [3171. 找到按位与最接近 K 的子数组](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README.md)
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<!-- tabs:start -->
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#### Python3

‎solution/1500-1599/1521.Find a Value of a Mysterious Function Closest to Target/README_EN.md

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The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Here, $n$ and $M$ are the length of the array $arr$ and the maximum value in the array $arr$, respectively.
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Similar problems:
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- [3171. Find Subarray With Bitwise AND Closest to K](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README_EN.md)
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<!-- tabs:start -->
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#### Python3

‎solution/3000-3099/3097.Shortest Subarray With OR at Least K II/README.md

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时间复杂度 $O(n \times \log M)$，空间复杂度 $O(\log M)$，其中 $n$ 和 $M$ 分别是数组的长度和数组中元素的最大值。
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相似题目：
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- [3171. 找到按位与最接近 K 的子数组](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README.md)
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<!-- tabs:start -->
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#### Python3

‎solution/3000-3099/3097.Shortest Subarray With OR at Least K II/README_EN.md

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The time complexity is $O(n \times \log M)$ and the space complexity is $O(\log M)$, where $n$ and $M$ are the length of the array and the maximum value of the elements in the array, respectively.
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Similar Problems:
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- [3171. Find Subarray With Bitwise AND Closest to K](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README_EN.md)
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<!-- tabs:start -->
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#### Python3

‎solution/3100-3199/3168.Minimum Number of Chairs in a Waiting Room/README.md

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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3168.Minimum%20Number%20of%20Chairs%20in%20a%20Waiting%20Room/README.md
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---
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<!-- problem:start -->
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# [3168. 候诊室中的最少椅子数](https://leetcode.cn/problems/minimum-number-of-chairs-in-a-waiting-room)
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[English Version](/solution/3100-3199/3168.Minimum%20Number%20of%20Chairs%20in%20a%20Waiting%20Room/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个字符串 <code>s</code>，模拟每秒钟的事件 <code>i</code>：</p>
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<ul>
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<li>如果 <code>s[i] == 'E'</code>，表示有一位顾客进入候诊室并占用一把椅子。</li>
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<li>如果 <code>s[i] == 'L'</code>，表示有一位顾客离开候诊室，从而释放一把椅子。</li>
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</ul>
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<p>返回保证每位进入候诊室的顾客都能有椅子坐的<strong> 最少 </strong>椅子数，假设候诊室最初是 <strong>空的 </strong>。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io">s = "EEEEEEE"</span></p>
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<p><strong>输出：</strong><span class="example-io">7</span></p>
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<p><strong>解释：</strong></p>
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<p>每秒后都有一个顾客进入候诊室，没有人离开。因此，至少需要 7 把椅子。</p>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io">s = "ELELEEL"</span></p>
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<p><strong>输出：</strong><span class="example-io">2</span></p>
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<p><strong>解释：</strong></p>
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<p>假设候诊室里有 2 把椅子。下表显示了每秒钟等候室的状态。</p>
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</div>
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<table>
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<tbody>
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<tr>
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<th>秒</th>
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<th>事件</th>
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<th>候诊室的人数</th>
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<th>可用的椅子数</th>
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</tr>
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<tr>
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<td>0</td>
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<td>Enter</td>
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<td>1</td>
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<td>1</td>
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</tr>
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<tr>
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<td>1</td>
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<td>Leave</td>
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<td>0</td>
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<td>2</td>
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</tr>
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<tr>
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<td>2</td>
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<td>Enter</td>
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<td>1</td>
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<td>1</td>
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</tr>
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<tr>
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<td>3</td>
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<td>Leave</td>
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<td>0</td>
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<td>2</td>
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</tr>
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<tr>
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<td>4</td>
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<td>Enter</td>
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<td>1</td>
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<td>1</td>
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</tr>
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<tr>
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<td>5</td>
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<td>Enter</td>
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<td>2</td>
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<td>0</td>
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</tr>
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<tr>
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<td>6</td>
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<td>Leave</td>
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<td>1</td>
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<td>1</td>
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</tr>
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</tbody>
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</table>
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<p><strong class="example">示例 3：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io">s = "ELEELEELLL"</span></p>
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<p><strong>输出：</strong><span class="example-io">3</span></p>
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<p><strong>解释：</strong></p>
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<p>假设候诊室里有 3 把椅子。下表显示了每秒钟等候室的状态。</p>
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</div>
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<table>
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<tbody>
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<tr>
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<th>秒</th>
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<th>事件</th>
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<th>候诊室的人数</th>
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<th>可用的椅子数</th>
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</tr>
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<tr>
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<td>0</td>
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<td>Enter</td>
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<td>1</td>
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<td>2</td>
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</tr>
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<tr>
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<td>1</td>
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<td>Leave</td>
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<td>0</td>
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<td>3</td>
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</tr>
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<tr>
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<td>2</td>
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<td>Enter</td>
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<td>1</td>
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<td>2</td>
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</tr>
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<tr>
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<td>3</td>
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<td>Enter</td>
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<td>2</td>
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<td>1</td>
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</tr>
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<tr>
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<td>4</td>
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<td>Leave</td>
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<td>1</td>
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<td>2</td>
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</tr>
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<tr>
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<td>5</td>
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<td>Enter</td>
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<td>2</td>
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<td>1</td>
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</tr>
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<tr>
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<td>6</td>
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<td>Enter</td>
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<td>3</td>
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<td>0</td>
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</tr>
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<tr>
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<td>7</td>
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<td>Leave</td>
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<td>2</td>
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<td>1</td>
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</tr>
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<tr>
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<td>8</td>
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<td>Leave</td>
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<td>1</td>
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<td>2</td>
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</tr>
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<tr>
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<td>9</td>
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<td>Leave</td>
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<td>0</td>
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<td>3</td>
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</tr>
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</tbody>
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</table>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 50</code></li>
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<li><code>s</code> 仅由字母 <code>'E'</code> 和 <code>'L'</code> 组成。</li>
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<li><code>s</code> 表示一个有效的进出序列。</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：模拟
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我们用变量 $\text{cnt}$ 来记录当前需要的椅子数，用变量 $\text{left}$ 来记录当前剩余的空椅子数。遍历字符串 $\text{s}$，如果当前字符是 'E'，那么如果有剩余的空椅子，就直接使用一个空椅子，否则需要增加一个椅子；如果当前字符是 'L'，那么剩余的空椅子数加一。
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遍历结束后，返回 $\text{cnt}$ 即可。
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时间复杂度 $O(n)$，其中 $n$ 为字符串 $\text{s}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def minimumChairs(self, s: str) -> int:
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cnt = left = 0
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for c in s:
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if c == "E":
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if left:
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left -= 1
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else:
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cnt += 1
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else:
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left += 1
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return cnt
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```
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#### Java
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```java
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class Solution {
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public int minimumChairs(String s) {
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int cnt = 0, left = 0;
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == 'E') {
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if (left > 0) {
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--left;
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} else {
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++cnt;
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}
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} else {
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++left;
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}
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}
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return cnt;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int minimumChairs(string s) {
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int cnt = 0, left = 0;
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for (char& c : s) {
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if (c == 'E') {
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if (left > 0) {
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--left;
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} else {
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++cnt;
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}
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} else {
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++left;
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}
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}
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return cnt;
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}
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};
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```
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#### Go
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```go
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func minimumChairs(s string) int {
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cnt, left := 0, 0
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for _, c := range s {
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if c == 'E' {
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if left > 0 {
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left--
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} else {
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cnt++
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}
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} else {
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left++
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}
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}
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return cnt
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}
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```
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#### TypeScript
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```ts
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function minimumChairs(s: string): number {
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let [cnt, left] = [0, 0];
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for (const c of s) {
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if (c === 'E') {
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if (left > 0) {
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--left;
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} else {
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++cnt;
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}
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} else {
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++left;
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}
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}
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return cnt;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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