feat: add solution to lc problem: No.3328

.prettierignore

@@ -25,3 +25,4 @@ node_modules/
 /solution/2200-2299/2253.Dynamic Unpivoting of a Table/Solution.sql
 /solution/3100-3199/3150.Invalid Tweets II/Solution.sql
 /solution/3100-3199/3198.Find Cities in Each State/Solution.sql
+/solution/3300-3399/3328.Find Cities in Each State II/Solution.sql

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

@@ -0,0 +1,193 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README.md
+tags:
+    - 数据库
+---
+
+<!-- problem:start -->
+
+# [3328. Find Cities in Each State II 🔒](https://leetcode.cn/problems/find-cities-in-each-state-ii)
+
+[English Version](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>Table: <code>cities</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| state       | varchar |
+| city        | varchar |
++-------------+---------+
+(state, city) is the 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</strong> in <strong>each state</strong> and analyze them based on the following requirements:</p>
+
+<ul>
+	<li>Combine all cities into a <strong>comma-separated</strong> string for each state.</li>
+	<li>Only include states that have <strong>at least</strong> <code>3</code> cities.</li>
+	<li>Only include states where <strong>at least one city</strong> starts with the <strong>same letter as the state name</strong>.</li>
+</ul>
+
+<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>
+
+<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          |
++--------------+---------------+
+| New York     | New York City |
+| New York     | Newark        |
+| New York     | Buffalo       |
+| New York     | Rochester     |
+| California   | San Francisco |
+| California   | Sacramento    |
+| California   | San Diego     |
+| California   | Los Angeles   |
+| Texas        | Tyler         |
+| Texas        | Temple        |
+| Texas        | Taylor        |
+| Texas        | Dallas        |
+| Pennsylvania | Philadelphia  |
+| Pennsylvania | Pittsburgh    |
+| Pennsylvania | Pottstown     |
++--------------+---------------+
+</pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++-------------+-------------------------------------------+-----------------------+
+| state       | cities                                    | matching_letter_count |
++-------------+-------------------------------------------+-----------------------+
+| Pennsylvania| Philadelphia, Pittsburgh, Pottstown       | 3                     |
+| Texas       | Dallas, Taylor, Temple, Tyler             | 2                     |
+| New York    | Buffalo, Newark, New York City, Rochester | 2                     |
++-------------+-------------------------------------------+-----------------------+
+</pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><strong>Pennsylvania</strong>:
+
+    <ul>
+    	<li>Has 3 cities (meets minimum requirement)</li>
+    	<li>All 3 cities start with &#39;P&#39; (same as state)</li>
+    	<li>matching_letter_count = 3</li>
+    </ul>
+    </li>
+    <li><strong>Texas</strong>:
+    <ul>
+    	<li>Has 4 cities (meets minimum requirement)</li>
+    	<li>2 cities (Temple, Tyler) start with &#39;T&#39; (same as state)</li>
+    	<li>matching_letter_count = 2</li>
+    </ul>
+    </li>
+    <li><strong>New York</strong>:
+    <ul>
+    	<li>Has 4 cities (meets minimum requirement)</li>
+    	<li>2 cities (Newark, New York City) start with &#39;N&#39; (same as state)</li>
+    	<li>matching_letter_count = 2</li>
+    </ul>
+    </li>
+    <li><strong>California</strong> is not included in the output because:
+    <ul>
+    	<li>Although it has 4 cities (meets minimum requirement)</li>
+    	<li>No cities start with &#39;C&#39; (doesn&#39;t meet the matching letter requirement)</li>
+    </ul>
+    </li>
+
+</ul>
+
+<p><strong>Note:</strong></p>
+
+<ul>
+	<li>Results are ordered by matching_letter_count in descending order</li>
+	<li>When matching_letter_count is the same (Texas and New York both have 2), they are ordered by state name alphabetically</li>
+	<li>Cities in each row are ordered alphabetically</li>
+</ul>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：分组聚合 + 过滤
+
+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。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    state,
+    GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') AS cities,
+    COUNT(
+        CASE
+            WHEN LEFT(city, 1) = LEFT(state, 1) THEN 1
+        END
+    ) AS matching_letter_count
+FROM cities
+GROUP BY 1
+HAVING COUNT(city) >= 3 AND matching_letter_count > 0
+ORDER BY 3 DESC, 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def state_city_analysis(cities: pd.DataFrame) -> pd.DataFrame:
+    cities["matching_letter"] = cities["city"].str[0] == cities["state"].str[0]
+
+    result = (
+        cities.groupby("state")
+        .agg(
+            cities=("city", lambda x: ", ".join(sorted(x))),
+            matching_letter_count=("matching_letter", "sum"),
+            city_count=("city", "count"),
+        )
+        .reset_index()
+    )
+
+    result = result[(result["city_count"] >= 3) & (result["matching_letter_count"] > 0)]
+
+    result = result.sort_values(
+        by=["matching_letter_count", "state"], ascending=[False, True]
+    )
+
+    result = result.drop(columns=["city_count"])
+
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

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

@@ -0,0 +1,193 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md
+tags:
+    - Database
+---
+
+<!-- problem:start -->
+
+# [3328. Find Cities in Each State II 🔒](https://leetcode.com/problems/find-cities-in-each-state-ii)
+
+[中文文档](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Table: <code>cities</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| state       | varchar |
+| city        | varchar |
++-------------+---------+
+(state, city) is the 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</strong> in <strong>each state</strong> and analyze them based on the following requirements:</p>
+
+<ul>
+	<li>Combine all cities into a <strong>comma-separated</strong> string for each state.</li>
+	<li>Only include states that have <strong>at least</strong> <code>3</code> cities.</li>
+	<li>Only include states where <strong>at least one city</strong> starts with the <strong>same letter as the state name</strong>.</li>
+</ul>
+
+<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>
+
+<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          |
++--------------+---------------+
+| New York     | New York City |
+| New York     | Newark        |
+| New York     | Buffalo       |
+| New York     | Rochester     |
+| California   | San Francisco |
+| California   | Sacramento    |
+| California   | San Diego     |
+| California   | Los Angeles   |
+| Texas        | Tyler         |
+| Texas        | Temple        |
+| Texas        | Taylor        |
+| Texas        | Dallas        |
+| Pennsylvania | Philadelphia  |
+| Pennsylvania | Pittsburgh    |
+| Pennsylvania | Pottstown     |
++--------------+---------------+
+</pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++-------------+-------------------------------------------+-----------------------+
+| state       | cities                                    | matching_letter_count |
++-------------+-------------------------------------------+-----------------------+
+| Pennsylvania| Philadelphia, Pittsburgh, Pottstown       | 3                     |
+| Texas       | Dallas, Taylor, Temple, Tyler             | 2                     |
+| New York    | Buffalo, Newark, New York City, Rochester | 2                     |
++-------------+-------------------------------------------+-----------------------+
+</pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><strong>Pennsylvania</strong>:
+
+    <ul>
+    	<li>Has 3 cities (meets minimum requirement)</li>
+    	<li>All 3 cities start with &#39;P&#39; (same as state)</li>
+    	<li>matching_letter_count = 3</li>
+    </ul>
+    </li>
+    <li><strong>Texas</strong>:
+    <ul>
+    	<li>Has 4 cities (meets minimum requirement)</li>
+    	<li>2 cities (Temple, Tyler) start with &#39;T&#39; (same as state)</li>
+    	<li>matching_letter_count = 2</li>
+    </ul>
+    </li>
+    <li><strong>New York</strong>:
+    <ul>
+    	<li>Has 4 cities (meets minimum requirement)</li>
+    	<li>2 cities (Newark, New York City) start with &#39;N&#39; (same as state)</li>
+    	<li>matching_letter_count = 2</li>
+    </ul>
+    </li>
+    <li><strong>California</strong> is not included in the output because:
+    <ul>
+    	<li>Although it has 4 cities (meets minimum requirement)</li>
+    	<li>No cities start with &#39;C&#39; (doesn&#39;t meet the matching letter requirement)</li>
+    </ul>
+    </li>
+
+</ul>
+
+<p><strong>Note:</strong></p>
+
+<ul>
+	<li>Results are ordered by matching_letter_count in descending order</li>
+	<li>When matching_letter_count is the same (Texas and New York both have 2), they are ordered by state name alphabetically</li>
+	<li>Cities in each row are ordered alphabetically</li>
+</ul>
+</div>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Group Aggregation + Filtering
+
+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.
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    state,
+    GROUP_CONCAT(city ORDER BY city SEPARATOR ', ') AS cities,
+    COUNT(
+        CASE
+            WHEN LEFT(city, 1) = LEFT(state, 1) THEN 1
+        END
+    ) AS matching_letter_count
+FROM cities
+GROUP BY 1
+HAVING COUNT(city) >= 3 AND matching_letter_count > 0
+ORDER BY 3 DESC, 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def state_city_analysis(cities: pd.DataFrame) -> pd.DataFrame:
+    cities["matching_letter"] = cities["city"].str[0] == cities["state"].str[0]
+
+    result = (
+        cities.groupby("state")
+        .agg(
+            cities=("city", lambda x: ", ".join(sorted(x))),
+            matching_letter_count=("matching_letter", "sum"),
+            city_count=("city", "count"),
+        )
+        .reset_index()
+    )
+
+    result = result[(result["city_count"] >= 3) & (result["matching_letter_count"] > 0)]
+
+    result = result.sort_values(
+        by=["matching_letter_count", "state"], ascending=[False, True]
+    )
+
+    result = result.drop(columns=["city_count"])
+
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->