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feat: add solutions to lc problem: No.3198 (#3164)
No.3198.Find Cities in Each State
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‎.prettierignore

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

‎solution/3100-3199/3198.Find Cities in Each State/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/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md
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---
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<!-- problem:start -->
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# [3198. 查找每个州的城市 🔒](https://leetcode.cn/problems/find-cities-in-each-state)
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[English Version](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>表：<code>cities</code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| state | varchar |
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| city | varchar |
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+-------------+---------+
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(state, city) 是这张表的主键（有不同值的列的组合）。
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这张表的每一行包含州名和其中的城市名。
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</pre>
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<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>
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<p>返回结果表以&nbsp;<code>state</code> <strong>升序&nbsp;</strong>排序。</p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong></p>
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<p>cities 表：</p>
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<pre class="example-io">
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+-------------+---------------+
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| state | city |
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+-------------+---------------+
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| California | Los Angeles |
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| California | San Francisco |
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| California | San Diego |
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| Texas | Houston |
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| Texas | Austin |
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| Texas | Dallas |
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| New York | New York City |
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| New York | Buffalo |
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| New York | Rochester |
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+-------------+---------------+
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</pre>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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+-------------+---------------------------------------+
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| state | cities |
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+-------------+---------------------------------------+
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| California | Los Angeles, San Diego, San Francisco |
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| New York | Buffalo, New York City, Rochester |
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| Texas | Austin, Dallas, Houston |
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+-------------+---------------------------------------+
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</pre>
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<p><strong>解释：</strong></p>
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<ul>
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<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>
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<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>
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<li><strong>Texas：</strong>所有城市 ("Austin", "Dallas", "Houston") 以逗号分隔的字符串列出。</li>
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</ul>
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<p><b>注意：</b>输出表以州名升序排序。</p>
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</div>
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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：分组聚合
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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。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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state,
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GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') cities
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FROM cities
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GROUP BY 1
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ORDER BY 1;
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```
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#### Pandas
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```python
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import pandas as pd
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def find_cities(cities: pd.DataFrame) -> pd.DataFrame:
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result = (
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cities.groupby("state")["city"]
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.apply(lambda x: ", ".join(sorted(x)))
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.reset_index()
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)
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result.columns = ["state", "cities"]
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return result
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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

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

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

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

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

‎solution/DATABASE_README.md

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

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| 3172 | [Second Day Verification](/solution/3100-3199/3172.Second%20Day%20Verification/README_EN.md) | `Database` | Easy | 🔒 |
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| 3182 | [Find Top Scoring Students](/solution/3100-3199/3182.Find%20Top%20Scoring%20Students/README_EN.md) | `Database` | Medium | 🔒 |
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| 3188 | [Find Top Scoring Students II](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md) | | Hard | 🔒 |
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| 3198 | [Find Cities in Each State](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md) | | Easy | 🔒 |
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## Copyright
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‎solution/README.md

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| 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 |
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| 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 |
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| 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 |
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| 3198 | [查找每个州的城市](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README.md) | | 简单 | 🔒 |
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## 版权
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‎solution/README_EN.md

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| 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 |
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| 3196 | [Maximize Total Cost of Alternating Subarrays](/solution/3100-3199/3196.Maximize%20Total%20Cost%20of%20Alternating%20Subarrays/README_EN.md) | | Medium | Weekly Contest 403 |
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| 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 |
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| 3198 | [Find Cities in Each State](/solution/3100-3199/3198.Find%20Cities%20in%20Each%20State/README_EN.md) | | Easy | 🔒 |
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## Copyright
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