feat: add weekly contest 400

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

@@ -75,6 +75,10 @@ tags:
 
 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。
 
+相似题目：
+
+-   [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)
+
 <!-- tabs:start -->
 
 #### Python3

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

@@ -76,6 +76,10 @@ If we fix the right endpoint $r$ each time, then the range of the left endpoint
 
 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.
 
+Similar problems:
+
+-   [3171. Find Subarray With Bitwise AND Closest to K](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README_EN.md)
+
 <!-- tabs:start -->
 
 #### Python3

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

@@ -92,6 +92,10 @@ tags:
 
 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。
 
+相似题目：
+
+-   [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)
+
 <!-- tabs:start -->
 
 #### Python3

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

@@ -90,6 +90,10 @@ Finally, we return the minimum length. If there is no subarray that meets the co
 
 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.
 
+Similar Problems:
+
+-   [3171. Find Subarray With Bitwise AND Closest to K](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20AND%20Closest%20to%20K/README_EN.md)
+
 <!-- tabs:start -->
 
 #### Python3

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

@@ -0,0 +1,318 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3168.Minimum%20Number%20of%20Chairs%20in%20a%20Waiting%20Room/README.md
+---
+
+<!-- problem:start -->
+
+# [3168. 候诊室中的最少椅子数](https://leetcode.cn/problems/minimum-number-of-chairs-in-a-waiting-room)
+
+[English Version](/solution/3100-3199/3168.Minimum%20Number%20of%20Chairs%20in%20a%20Waiting%20Room/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个字符串 <code>s</code>，模拟每秒钟的事件 <code>i</code>：</p>
+
+<ul>
+	<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>
+	<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>
+</ul>
+
+<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>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">s = "EEEEEEE"</span></p>
+
+<p><strong>输出：</strong><span class="example-io">7</span></p>
+
+<p><strong>解释：</strong></p>
+
+<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>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">s = "ELELEEL"</span></p>
+
+<p><strong>输出：</strong><span class="example-io">2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<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>
+</div>
+<table>
+	<tbody>
+		<tr>
+			<th>秒</th>
+			<th>事件</th>
+			<th>候诊室的人数</th>
+			<th>可用的椅子数</th>
+		</tr>
+		<tr>
+			<td>0</td>
+			<td>Enter</td>
+			<td>1</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>1</td>
+			<td>Leave</td>
+			<td>0</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>2</td>
+			<td>Enter</td>
+			<td>1</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>3</td>
+			<td>Leave</td>
+			<td>0</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>4</td>
+			<td>Enter</td>
+			<td>1</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>5</td>
+			<td>Enter</td>
+			<td>2</td>
+			<td>0</td>
+		</tr>
+		<tr>
+			<td>6</td>
+			<td>Leave</td>
+			<td>1</td>
+			<td>1</td>
+		</tr>
+	</tbody>
+</table>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">s = "ELEELEELLL"</span></p>
+
+<p><strong>输出：</strong><span class="example-io">3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<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>
+</div>
+<table>
+	<tbody>
+		<tr>
+			<th>秒</th>
+			<th>事件</th>
+			<th>候诊室的人数</th>
+			<th>可用的椅子数</th>
+		</tr>
+		<tr>
+			<td>0</td>
+			<td>Enter</td>
+			<td>1</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>1</td>
+			<td>Leave</td>
+			<td>0</td>
+			<td>3</td>
+		</tr>
+		<tr>
+			<td>2</td>
+			<td>Enter</td>
+			<td>1</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>3</td>
+			<td>Enter</td>
+			<td>2</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>4</td>
+			<td>Leave</td>
+			<td>1</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>5</td>
+			<td>Enter</td>
+			<td>2</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>6</td>
+			<td>Enter</td>
+			<td>3</td>
+			<td>0</td>
+		</tr>
+		<tr>
+			<td>7</td>
+			<td>Leave</td>
+			<td>2</td>
+			<td>1</td>
+		</tr>
+		<tr>
+			<td>8</td>
+			<td>Leave</td>
+			<td>1</td>
+			<td>2</td>
+		</tr>
+		<tr>
+			<td>9</td>
+			<td>Leave</td>
+			<td>0</td>
+			<td>3</td>
+		</tr>
+	</tbody>
+</table>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 50</code></li>
+	<li><code>s</code> 仅由字母 <code>'E'</code> 和 <code>'L'</code> 组成。</li>
+	<li><code>s</code> 表示一个有效的进出序列。</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+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。
+
+遍历结束后，返回 $\text{cnt}$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为字符串 $\text{s}$ 的长度。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minimumChairs(self, s: str) -> int:
+        cnt = left = 0
+        for c in s:
+            if c == "E":
+                if left:
+                    left -= 1
+                else:
+                    cnt += 1
+            else:
+                left += 1
+        return cnt
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minimumChairs(String s) {
+        int cnt = 0, left = 0;
+        for (int i = 0; i < s.length(); ++i) {
+            if (s.charAt(i) == 'E') {
+                if (left > 0) {
+                    --left;
+                } else {
+                    ++cnt;
+                }
+            } else {
+                ++left;
+            }
+        }
+        return cnt;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minimumChairs(string s) {
+        int cnt = 0, left = 0;
+        for (char& c : s) {
+            if (c == 'E') {
+                if (left > 0) {
+                    --left;
+                } else {
+                    ++cnt;
+                }
+            } else {
+                ++left;
+            }
+        }
+        return cnt;
+    }
+};
+```
+
+#### Go
+
+```go
+func minimumChairs(s string) int {
+	cnt, left := 0, 0
+	for _, c := range s {
+		if c == 'E' {
+			if left > 0 {
+				left--
+			} else {
+				cnt++
+			}
+		} else {
+			left++
+		}
+	}
+	return cnt
+}
+```
+
+#### TypeScript
+
+```ts
+function minimumChairs(s: string): number {
+    let [cnt, left] = [0, 0];
+    for (const c of s) {
+        if (c === 'E') {
+            if (left > 0) {
+                --left;
+            } else {
+                ++cnt;
+            }
+        } else {
+            ++left;
+        }
+    }
+    return cnt;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->