feat: add solutions to lc problem: No.3198

.prettierignore

@@ -24,4 +24,5 @@ node_modules/
 /solution/2200-2299/2230.The Users That Are Eligible for Discount/Solution.sql
 /solution/2200-2299/2252.Dynamic Pivoting of a Table/Solution.sql
 /solution/2200-2299/2253.Dynamic Unpivoting of a Table/Solution.sql
-/solution/3100-3199/3150.Invalid Tweets II/Solution.sql
+/solution/3100-3199/3150.Invalid Tweets II/Solution.sql
+/solution/3100-3199/3198.Find Cities in Each State/Solution.sql

solution/3100-3199/3198.Find Cities in Each State/README.md

@@ -0,0 +1,128 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md
+---
+
+<!-- problem:start -->
+
+# [3198. 查找每个州的城市 🔒](https://leetcode.cn/problems/find-cities-in-each-state)
+
+[English Version](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>表：<code>cities</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    | 
++-------------+---------+
+| state       | varchar |
+| city        | varchar |
++-------------+---------+
+(state, city) 是这张表的主键（有不同值的列的组合）。
+这张表的每一行包含州名和其中的城市名。
+</pre>
+
+<p>编写一个解决方案来 <strong>查找每个州的所有城市</strong>，并将它们组合成 <strong>一个逗号分隔</strong> 的字符串。</p>
+
+<p>返回结果表以&nbsp;<code>state</code> <strong>升序&nbsp;</strong>排序。</p>
+
+<p>结果格式如下所示。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong></p>
+
+<p>cities 表：</p>
+
+<pre class="example-io">
++-------------+---------------+
+| state       | city          |
++-------------+---------------+
+| California  | Los Angeles   |
+| California  | San Francisco |
+| California  | San Diego     |
+| Texas       | Houston       |
+| Texas       | Austin        |
+| Texas       | Dallas        |
+| New York    | New York City |
+| New York    | Buffalo       |
+| New York    | Rochester     |
++-------------+---------------+
+</pre>
+
+<p><strong>输出：</strong></p>
+
+<pre class="example-io">
++-------------+---------------------------------------+
+| state       | cities                                |
++-------------+---------------------------------------+
+| California  | Los Angeles, San Diego, San Francisco |
+| New York    | Buffalo, New York City, Rochester     |
+| Texas       | Austin, Dallas, Houston               |
++-------------+---------------------------------------+
+</pre>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><strong>California：</strong>所有城市 ("Los Angeles", "San Diego", "San Francisco") 以逗号分隔的字符串列出。</li>
+	<li><strong>New York：</strong>所有城市 ("Buffalo", "New York City", "Rochester") 以逗号分隔的字符串列出。</li>
+	<li><strong>Texas：</strong>所有城市 ("Austin", "Dallas", "Houston") 以逗号分隔的字符串列出。</li>
+</ul>
+
+<p><b>注意：</b>输出表以州名升序排序。</p>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：分组聚合
+
+我们可以先按照 `state` 字段进行分组，然后对每个分组内的 `city` 字段进行排序，最后使用 `GROUP_CONCAT` 函数将排序后的城市名连接成一个逗号分隔的字符串。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    state,
+    GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') cities
+FROM cities
+GROUP BY 1
+ORDER BY 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def find_cities(cities: pd.DataFrame) -> pd.DataFrame:
+    result = (
+        cities.groupby("state")["city"]
+        .apply(lambda x: ", ".join(sorted(x)))
+        .reset_index()
+    )
+    result.columns = ["state", "cities"]
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3198.Find Cities in Each State/README_EN.md

@@ -0,0 +1,127 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3198. Find Cities in Each State 🔒](https://leetcode.com/problems/find-cities-in-each-state)
+
+[中文文档](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Table: <code>cities</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    | 
++-------------+---------+
+| state       | varchar |
+| city        | varchar |
++-------------+---------+
+(state, city) is the primary key (combination of columns with unique values) for this table.
+Each row of this table contains the state name and the city name within that state.
+</pre>
+
+<p>Write a solution to find <strong>all the cities in each state</strong> and combine them into a <strong>single comma-separated</strong> string.</p>
+
+<p>Return <em>the result table ordered by</em> <code>state</code> <em>in <strong>ascending</strong> order</em>.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p>cities table:</p>
+
+<pre class="example-io">
++-------------+---------------+
+| state       | city          |
++-------------+---------------+
+| California  | Los Angeles   |
+| California  | San Francisco |
+| California  | San Diego     |
+| Texas       | Houston       |
+| Texas       | Austin        |
+| Texas       | Dallas        |
+| New York    | New York City |
+| New York    | Buffalo       |
+| New York    | Rochester     |
++-------------+---------------+
+</pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++-------------+---------------------------------------+
+| state       | cities                                |
++-------------+---------------------------------------+
+| California  | Los Angeles, San Diego, San Francisco |
+| New York    | Buffalo, New York City, Rochester     |
+| Texas       | Austin, Dallas, Houston               |
++-------------+---------------------------------------+
+</pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><strong>California:</strong> All cities (&quot;Los Angeles&quot;, &quot;San Diego&quot;, &quot;San Francisco&quot;) are listed in a comma-separated string.</li>
+	<li><strong>New York:</strong> All cities (&quot;Buffalo&quot;, &quot;New York City&quot;, &quot;Rochester&quot;) are listed in a comma-separated string.</li>
+	<li><strong>Texas:</strong> All cities (&quot;Austin&quot;, &quot;Dallas&quot;, &quot;Houston&quot;) are listed in a comma-separated string.</li>
+</ul>
+
+<p><strong>Note:</strong> The output table is ordered by the state name in ascending order.</p>
+</div>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Grouping and Aggregation
+
+We can first group by the `state` field, then sort the `city` field within each group, and finally use the `GROUP_CONCAT` function to concatenate the sorted city names into a comma-separated string.
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    state,
+    GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') cities
+FROM cities
+GROUP BY 1
+ORDER BY 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def find_cities(cities: pd.DataFrame) -> pd.DataFrame:
+    result = (
+        cities.groupby("state")["city"]
+        .apply(lambda x: ", ".join(sorted(x)))
+        .reset_index()
+    )
+    result.columns = ["state", "cities"]
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3198.Find Cities in Each State/Solution.py

@@ -0,0 +1,11 @@
+import pandas as pd
+
+
+def find_cities(cities: pd.DataFrame) -> pd.DataFrame:
+    result = (
+        cities.groupby("state")["city"]
+        .apply(lambda x: ", ".join(sorted(x)))
+        .reset_index()
+    )
+    result.columns = ["state", "cities"]
+    return result

solution/3100-3199/3198.Find Cities in Each State/Solution.sql

@@ -0,0 +1,7 @@
+# Write your MySQL query statement below
+SELECT
+    state,
+    GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') cities
+FROM cities
+GROUP BY 1
+ORDER BY 1;

solution/DATABASE_README.md

@@ -284,6 +284,7 @@
 | 3172 | [第二天验证](/solution/3100-3199/3172.Second%20Day%20Verification/README.md)                                                                                 | `数据库` | 简单 | 🔒   |
 | 3182 | [查找得分最高的学生](/solution/3100-3199/3182.Find%20Top%20Scoring%20Students/README.md)                                                                     | `数据库` | 中等 | 🔒   |
 | 3188 | [查找得分最高的学生 II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)                                                             |          | 困难 | 🔒   |
+| 3198 | [查找每个州的城市](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md)                                                                     |          | 简单 | 🔒   |
 
 ## 版权
 

solution/DATABASE_README_EN.md

@@ -282,6 +282,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 | 3172 | [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md)                                                                                                 | `Database` | Easy       | 🔒     |
 | 3182 | [Find Top Scoring Students](/solution/3100-3199/3182.Find%20Top%20Scoring%20Students/README_EN.md)                                                                                           | `Database` | Medium     | 🔒     |
 | 3188 | [Find Top Scoring Students II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md)                                                                                   |            | Hard       | 🔒     |
+| 3198 | [Find Cities in Each State](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md)                                                                                         |            | Easy       | 🔒     |
 
 ## Copyright
 

solution/README.md

@@ -3208,6 +3208,7 @@
 |  3195  |  [包含所有 1 的最小矩形面积 I](/solution/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/README.md)  |    |  中等  |  第 403 场周赛  |
 |  3196  |  [最大化子数组的总成本](/solution/3100-3199/3196.Maximize%20Total%20Cost%20of%20Alternating%20Subarrays/README.md)  |    |  中等  |  第 403 场周赛  |
 |  3197  |  [包含所有 1 的最小矩形面积 II](/solution/3100-3199/3197.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20II/README.md)  |    |  困难  |  第 403 场周赛  |
+|  3198  |  [查找每个州的城市](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md)  |    |  简单  |  🔒  |
 
 ## 版权
 

solution/README_EN.md

@@ -3206,6 +3206,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3195  |  [Find the Minimum Area to Cover All Ones I](/solution/3100-3199/3195.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20I/README_EN.md)  |    |  Medium  |  Weekly Contest 403  |
 |  3196  |  [Maximize Total Cost of Alternating Subarrays](/solution/3100-3199/3196.Maximize%20Total%20Cost%20of%20Alternating%20Subarrays/README_EN.md)  |    |  Medium  |  Weekly Contest 403  |
 |  3197  |  [Find the Minimum Area to Cover All Ones II](/solution/3100-3199/3197.Find%20the%20Minimum%20Area%20to%20Cover%20All%20Ones%20II/README_EN.md)  |    |  Hard  |  Weekly Contest 403  |
+|  3198  |  [Find Cities in Each State](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md)  |    |  Easy  |  🔒  |
 
 ## Copyright