feat: add weekly contest 372 (#1983)

Author: acbin

solution/2900-2999/2937.Make Three Strings Equal/README.md

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+# [2937. 使三个字符串相等](https://leetcode.cn/problems/make-three-strings-equal)
+
+[English Version](/solution/2900-2999/2937.Make%20Three%20Strings%20Equal/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你三个字符串 <code>s1</code>、<code>s2</code> 和 <code>s3</code>。 你可以根据需要对这三个字符串执行以下操作 <strong>任意次数</strong> <!-- notionvc: b5178de7-3318-4129-b7d9-726b47e90621 -->。</p>
+
+<p>在每次操作中，你可以选择其中一个长度至少为 <code>2</code> 的字符串 <!-- notionvc: 3342ac46-33c8-4010-aacd-e58678ce31ef --> 并删除其 <strong>最右位置上</strong> 的字符。</p>
+
+<p>如果存在某种方法能够使这三个字符串相等，请返回使它们相等所需的 <strong>最小</strong> 操作次数；否则，返回 <code>-1</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s1 = "abc"，s2 = "abb"，s3 = "ab"
+<strong>输出：</strong>2
+<strong>解释：</strong>对 s1 和 s2 进行一次操作后，可以得到三个相等的字符串。
+可以证明，不可能用少于两次操作使它们相等。</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s1 = "dac"，s2 = "bac"，s3 = "cac"
+<strong>输出：</strong>-1
+<strong>解释：</strong>因为 s1 和 s2 的最左位置上的字母<!-- notionvc: 47239f7c-eec1-49f8-af79-c206ec88cb07 -->不相等，所以无论进行多少次操作，它们都不可能相等。因此答案是 -1 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s1.length, s2.length, s3.length &lt;= 100</code></li>
+	<li><code>s1</code>、<code>s2</code> 和 <code>s3</code> 仅由小写英文字母组成。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2937.Make Three Strings Equal/README_EN.md

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+# [2937. Make Three Strings Equal](https://leetcode.com/problems/make-three-strings-equal)
+
+[中文文档](/solution/2900-2999/2937.Make%20Three%20Strings%20Equal/README.md)
+
+## Description
+
+<p>You are given three strings <code>s1</code>, <code>s2</code>, and <code>s3</code>. You have to perform the following operation on these three strings <strong>as many times</strong> as you want<!-- notionvc: b5178de7-3318-4129-b7d9-726b47e90621 -->.</p>
+
+<p>In one operation you can choose one of these three strings such that its length is at least <code>2</code><!-- notionvc: 3342ac46-33c8-4010-aacd-e58678ce31ef --> and delete the <strong>rightmost</strong> character of it.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of operations you need to perform to make the three strings equal if there is a way to make them equal, otherwise, return </em><code>-1</code><em>.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;abc&quot;, s2 = &quot;abb&quot;, s3 = &quot;ab&quot;
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> Performing operations on s1 and s2 once will lead to three equal strings.
+It can be shown that there is no way to make them equal with less than two operations.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s1 = &quot;dac&quot;, s2 = &quot;bac&quot;, s3 = &quot;cac&quot;
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> Because the leftmost letters<!-- notionvc: 47239f7c-eec1-49f8-af79-c206ec88cb07 --> of s1 and s2 are not equal, they could not be equal after any number of operations. So the answer is -1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s1.length, s2.length, s3.length &lt;= 100</code></li>
+	<li><font face="monospace"><code>s1</code>,</font> <code><font face="monospace">s2</font></code><font face="monospace"> and</font> <code><font face="monospace">s3</font></code> consist only of lowercase English letters.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2938.Separate Black and White Balls/README.md

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+# [2938. 区分黑球与白球](https://leetcode.cn/problems/separate-black-and-white-balls)
+
+[English Version](/solution/2900-2999/2938.Separate%20Black%20and%20White%20Balls/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>桌子上有 <code>n</code> 个球，每个球的颜色不是黑色，就是白色。</p>
+
+<p>给你一个长度为 <code>n</code> 、下标从 <strong>0</strong> 开始的二进制字符串 <code>s</code>，其中 <code>1</code> 和 <code>0</code> 分别代表黑色和白色的球。</p>
+
+<p>在每一步中，你可以选择两个相邻的球并交换它们。</p>
+
+<p>返回「将所有黑色球都移到右侧，所有白色球都移到左侧所需的 <strong>最小步数</strong>」。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "101"
+<strong>输出：</strong>1
+<strong>解释：</strong>我们可以按以下方式将所有黑色球移到右侧：
+- 交换 s[0] 和 s[1]，s = "011"。
+最开始，1 没有都在右侧，需要至少 1 步将其移到右侧。</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "100"
+<strong>输出：</strong>2
+<strong>解释：</strong>我们可以按以下方式将所有黑色球移到右侧：
+- 交换 s[0] 和 s[1]，s = "010"。
+- 交换 s[1] 和 s[2]，s = "001"。
+可以证明所需的最小步数为 2 。
+</pre>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "0111"
+<strong>输出：</strong>0
+<strong>解释：</strong>所有黑色球都已经在右侧。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> 不是 <code>'0'</code>，就是 <code>'1'</code>。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2938.Separate Black and White Balls/README_EN.md

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+# [2938. Separate Black and White Balls](https://leetcode.com/problems/separate-black-and-white-balls)
+
+[中文文档](/solution/2900-2999/2938.Separate%20Black%20and%20White%20Balls/README.md)
+
+## Description
+
+<p>There are <code>n</code> balls on a table, each ball has a color black or white.</p>
+
+<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> of length <code>n</code>, where <code>1</code> and <code>0</code> represent black and white balls, respectively.</p>
+
+<p>In each step, you can choose two adjacent balls and swap them.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of steps to group all the black balls to the right and all the white balls to the left</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;101&quot;
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> We can group all the black balls to the right in the following way:
+- Swap s[0] and s[1], s = &quot;011&quot;.
+Initially, 1s are not grouped together, requiring at least 1 step to group them to the right.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;100&quot;
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> We can group all the black balls to the right in the following way:
+- Swap s[0] and s[1], s = &quot;010&quot;.
+- Swap s[1] and s[2], s = &quot;001&quot;.
+It can be proven that the minimum number of steps needed is 2.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;0111&quot;
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> All the black balls are already grouped to the right.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->