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feat: add solution to lc problem: No.3328 (#3656)
No.3328.Find Cities in Each State II
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‎.prettierignore

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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/3100-3199/3198.Find Cities in Each State/Solution.sql
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/solution/3300-3399/3328.Find Cities in Each State II/Solution.sql

‎solution/3300-3399/3328.Find Cities in Each State II/README.md

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README.md
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tags:
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- 数据库
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---
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<!-- problem:start -->
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# [3328. Find Cities in Each State II 🔒](https://leetcode.cn/problems/find-cities-in-each-state-ii)
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[English Version](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md)
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## 题目描述
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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 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</strong> in <strong>each state</strong> and analyze them based on the following requirements:</p>
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<ul>
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<li>Combine all cities into a <strong>comma-separated</strong> string for each state.</li>
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<li>Only include states that have <strong>at least</strong> <code>3</code> cities.</li>
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<li>Only include states where <strong>at least one city</strong> starts with the <strong>same letter as the state name</strong>.</li>
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</ul>
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<p>Return <em>the result table ordered by</em> <em>the count of matching-letter cities in <strong>descending</strong> order</em>&nbsp;<em>and then by state name 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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| New York | New York City |
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| New York | Newark |
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| New York | Buffalo |
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| New York | Rochester |
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| California | San Francisco |
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| California | Sacramento |
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| California | San Diego |
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| California | Los Angeles |
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| Texas | Tyler |
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| Texas | Temple |
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| Texas | Taylor |
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| Texas | Dallas |
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| Pennsylvania | Philadelphia |
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| Pennsylvania | Pittsburgh |
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| Pennsylvania | Pottstown |
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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 | matching_letter_count |
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+-------------+-------------------------------------------+-----------------------+
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| Pennsylvania| Philadelphia, Pittsburgh, Pottstown | 3 |
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| Texas | Dallas, Taylor, Temple, Tyler | 2 |
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| New York | Buffalo, Newark, New York City, Rochester | 2 |
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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>Pennsylvania</strong>:
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<ul>
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<li>Has 3 cities (meets minimum requirement)</li>
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<li>All 3 cities start with &#39;P&#39; (same as state)</li>
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<li>matching_letter_count = 3</li>
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</ul>
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</li>
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<li><strong>Texas</strong>:
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<ul>
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<li>Has 4 cities (meets minimum requirement)</li>
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<li>2 cities (Temple, Tyler) start with &#39;T&#39; (same as state)</li>
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<li>matching_letter_count = 2</li>
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</ul>
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</li>
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<li><strong>New York</strong>:
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<ul>
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<li>Has 4 cities (meets minimum requirement)</li>
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<li>2 cities (Newark, New York City) start with &#39;N&#39; (same as state)</li>
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<li>matching_letter_count = 2</li>
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</ul>
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</li>
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<li><strong>California</strong> is not included in the output because:
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<ul>
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<li>Although it has 4 cities (meets minimum requirement)</li>
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<li>No cities start with &#39;C&#39; (doesn&#39;t meet the matching letter requirement)</li>
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</ul>
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</li>
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</ul>
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<p><strong>Note:</strong></p>
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<ul>
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<li>Results are ordered by matching_letter_count in descending order</li>
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<li>When matching_letter_count is the same (Texas and New York both have 2), they are ordered by state name alphabetically</li>
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<li>Cities in each row are ordered alphabetically</li>
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</ul>
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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 + 过滤
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我们可以将 `cities` 表按照 `state` 字段进行分组聚合，然后对每个分组进行过滤，筛选出满足条件的分组。
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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 ', ') AS cities,
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COUNT(
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CASE
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WHEN LEFT(city, 1) = LEFT(state, 1) THEN 1
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END
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) AS matching_letter_count
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FROM cities
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GROUP BY 1
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HAVING COUNT(city) >= 3 AND matching_letter_count > 0
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ORDER BY 3 DESC, 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 state_city_analysis(cities: pd.DataFrame) -> pd.DataFrame:
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cities["matching_letter"] = cities["city"].str[0] == cities["state"].str[0]
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result = (
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cities.groupby("state")
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.agg(
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cities=("city", lambda x: ", ".join(sorted(x))),
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matching_letter_count=("matching_letter", "sum"),
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city_count=("city", "count"),
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)
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.reset_index()
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)
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result = result[(result["city_count"] >= 3) & (result["matching_letter_count"] > 0)]
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result = result.sort_values(
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by=["matching_letter_count", "state"], ascending=[False, True]
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)
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result = result.drop(columns=["city_count"])
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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/3300-3399/3328.Find Cities in Each State II/README_EN.md

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---
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comments: true
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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md
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tags:
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- Database
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---
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<!-- problem:start -->
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# [3328. Find Cities in Each State II 🔒](https://leetcode.com/problems/find-cities-in-each-state-ii)
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[中文文档](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/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 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</strong> in <strong>each state</strong> and analyze them based on the following requirements:</p>
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<ul>
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<li>Combine all cities into a <strong>comma-separated</strong> string for each state.</li>
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<li>Only include states that have <strong>at least</strong> <code>3</code> cities.</li>
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<li>Only include states where <strong>at least one city</strong> starts with the <strong>same letter as the state name</strong>.</li>
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</ul>
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<p>Return <em>the result table ordered by</em> <em>the count of matching-letter cities in <strong>descending</strong> order</em>&nbsp;<em>and then by state name 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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| New York | New York City |
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| New York | Newark |
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| New York | Buffalo |
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| New York | Rochester |
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| California | San Francisco |
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| California | Sacramento |
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| California | San Diego |
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| California | Los Angeles |
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| Texas | Tyler |
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| Texas | Temple |
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| Texas | Taylor |
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| Texas | Dallas |
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| Pennsylvania | Philadelphia |
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| Pennsylvania | Pittsburgh |
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| Pennsylvania | Pottstown |
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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 | matching_letter_count |
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+-------------+-------------------------------------------+-----------------------+
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| Pennsylvania| Philadelphia, Pittsburgh, Pottstown | 3 |
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| Texas | Dallas, Taylor, Temple, Tyler | 2 |
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| New York | Buffalo, Newark, New York City, Rochester | 2 |
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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>Pennsylvania</strong>:
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<ul>
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<li>Has 3 cities (meets minimum requirement)</li>
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<li>All 3 cities start with &#39;P&#39; (same as state)</li>
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<li>matching_letter_count = 3</li>
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</ul>
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</li>
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<li><strong>Texas</strong>:
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<ul>
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<li>Has 4 cities (meets minimum requirement)</li>
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<li>2 cities (Temple, Tyler) start with &#39;T&#39; (same as state)</li>
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<li>matching_letter_count = 2</li>
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</ul>
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</li>
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<li><strong>New York</strong>:
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<ul>
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<li>Has 4 cities (meets minimum requirement)</li>
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<li>2 cities (Newark, New York City) start with &#39;N&#39; (same as state)</li>
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<li>matching_letter_count = 2</li>
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</ul>
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</li>
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<li><strong>California</strong> is not included in the output because:
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<ul>
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<li>Although it has 4 cities (meets minimum requirement)</li>
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<li>No cities start with &#39;C&#39; (doesn&#39;t meet the matching letter requirement)</li>
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</ul>
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</li>
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</ul>
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<p><strong>Note:</strong></p>
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<ul>
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<li>Results are ordered by matching_letter_count in descending order</li>
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<li>When matching_letter_count is the same (Texas and New York both have 2), they are ordered by state name alphabetically</li>
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<li>Cities in each row are ordered alphabetically</li>
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</ul>
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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: Group Aggregation + Filtering
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We can group the `cities` table by the `state` field, then apply filtering on each group to retain only the groups that meet the specified conditions.
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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 ', ') AS cities,
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COUNT(
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CASE
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WHEN LEFT(city, 1) = LEFT(state, 1) THEN 1
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END
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) AS matching_letter_count
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FROM cities
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GROUP BY 1
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HAVING COUNT(city) >= 3 AND matching_letter_count > 0
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ORDER BY 3 DESC, 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 state_city_analysis(cities: pd.DataFrame) -> pd.DataFrame:
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cities["matching_letter"] = cities["city"].str[0] == cities["state"].str[0]
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result = (
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cities.groupby("state")
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.agg(
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cities=("city", lambda x: ", ".join(sorted(x))),
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matching_letter_count=("matching_letter", "sum"),
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city_count=("city", "count"),
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)
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.reset_index()
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)
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result = result[(result["city_count"] >= 3) & (result["matching_letter_count"] > 0)]
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result = result.sort_values(
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by=["matching_letter_count", "state"], ascending=[False, True]
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)
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result = result.drop(columns=["city_count"])
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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 -->

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