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163 changes: 163 additions & 0 deletions solution/3300-3399/3338.Second Highest Salary II/README.md
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---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md
tags:
- 数据库
---

<!-- problem:start -->

# [3338. Second Highest Salary II 🔒](https://leetcode.cn/problems/second-highest-salary-ii)

[English Version](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md)

## 题目描述

<!-- description:start -->

<p>Table: <code>employees</code></p>

<pre>
+------------------+---------+
| Column Name | Type |
+------------------+---------+
| emp_id | int |
| salary | int |
| dept | varchar |
+------------------+---------+
emp_id is the unique key for this table.
Each row of this table contains information about an employee including their ID, name, manager, salary, department, start date, and building assignment.
</pre>

<p>Write a solution to find the employees who earn the <strong>second-highest salary</strong> in each department. If <strong>multiple employees have the second-highest salary</strong>, <strong>include</strong> <strong>all employees</strong> with <strong>that salary</strong>.</p>

<p>Return <em>the result table</em> <em>ordered by</em> <code>emp_id</code> <em>in</em> <em><strong>ascending</strong></em> <em>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>employees table:</p>

<pre class="example-io">
+--------+--------+-----------+
| emp_id | salary | dept |
+--------+--------+-----------+
| 1 | 70000 | Sales |
| 2 | 80000 | Sales |
| 3 | 80000 | Sales |
| 4 | 90000 | Sales |
| 5 | 55000 | IT |
| 6 | 65000 | IT |
| 7 | 65000 | IT |
| 8 | 50000 | Marketing |
| 9 | 55000 | Marketing |
| 10 | 55000 | HR |
+--------+--------+-----------+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+--------+-----------+
| emp_id | dept |
+--------+-----------+
| 2 | Sales |
| 3 | Sales |
| 5 | IT |
| 8 | Marketing |
+--------+-----------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li><strong>Sales Department</strong>:

<ul>
<li>Highest salary is 90000 (emp_id: 4)</li>
<li>Second-highest salary is 80000 (emp_id: 2, 3)</li>
<li>Both employees with salary 80000 are included</li>
</ul>
</li>
<li><strong>IT Department</strong>:
<ul>
<li>Highest salary is 65000 (emp_id: 6, 7)</li>
<li>Second-highest salary is 55000 (emp_id: 5)</li>
<li>Only emp_id 5 is included as they have the second-highest salary</li>
</ul>
</li>
<li><strong>Marketing Department</strong>:
<ul>
<li>Highest salary is 55000 (emp_id: 9)</li>
<li>Second-highest salary is 50000 (emp_id: 8)</li>
<li>Employee 8&nbsp;is included</li>
</ul>
</li>
<li><strong>HR Department</strong>:
<ul>
<li>Only has one employee</li>
<li>Not included in the result as it has fewer than 2 employees</li>
</ul>
</li>

</ul>
</div>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：窗口函数

我们可以使用 `DENSE_RANK()` 窗口函数来为每个部门的员工按照工资降序排名，然后筛选出排名为 $2$ 的员工即可。

<!-- tabs:start -->

#### MySQL

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT
emp_id,
dept,
DENSE_RANK() OVER (
PARTITION BY dept
ORDER BY salary DESC
) rk
FROM Employees
)
SELECT emp_id, dept
FROM T
WHERE rk = 2
ORDER BY 1;
```

#### Pandas

```python
import pandas as pd


def find_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
employees["rk"] = employees.groupby("dept")["salary"].rank(
method="dense", ascending=False
)
second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
return second_highest.sort_values(by="emp_id")
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
163 changes: 163 additions & 0 deletions solution/3300-3399/3338.Second Highest Salary II/README_EN.md
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---
comments: true
difficulty: Medium
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md
tags:
- Database
---

<!-- problem:start -->

# [3338. Second Highest Salary II 🔒](https://leetcode.com/problems/second-highest-salary-ii)

[中文文档](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md)

## Description

<!-- description:start -->

<p>Table: <code>employees</code></p>

<pre>
+------------------+---------+
| Column Name | Type |
+------------------+---------+
| emp_id | int |
| salary | int |
| dept | varchar |
+------------------+---------+
emp_id is the unique key for this table.
Each row of this table contains information about an employee including their ID, name, manager, salary, department, start date, and building assignment.
</pre>

<p>Write a solution to find the employees who earn the <strong>second-highest salary</strong> in each department. If <strong>multiple employees have the second-highest salary</strong>, <strong>include</strong> <strong>all employees</strong> with <strong>that salary</strong>.</p>

<p>Return <em>the result table</em> <em>ordered by</em> <code>emp_id</code> <em>in</em> <em><strong>ascending</strong></em> <em>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>employees table:</p>

<pre class="example-io">
+--------+--------+-----------+
| emp_id | salary | dept |
+--------+--------+-----------+
| 1 | 70000 | Sales |
| 2 | 80000 | Sales |
| 3 | 80000 | Sales |
| 4 | 90000 | Sales |
| 5 | 55000 | IT |
| 6 | 65000 | IT |
| 7 | 65000 | IT |
| 8 | 50000 | Marketing |
| 9 | 55000 | Marketing |
| 10 | 55000 | HR |
+--------+--------+-----------+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+--------+-----------+
| emp_id | dept |
+--------+-----------+
| 2 | Sales |
| 3 | Sales |
| 5 | IT |
| 8 | Marketing |
+--------+-----------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li><strong>Sales Department</strong>:

<ul>
<li>Highest salary is 90000 (emp_id: 4)</li>
<li>Second-highest salary is 80000 (emp_id: 2, 3)</li>
<li>Both employees with salary 80000 are included</li>
</ul>
</li>
<li><strong>IT Department</strong>:
<ul>
<li>Highest salary is 65000 (emp_id: 6, 7)</li>
<li>Second-highest salary is 55000 (emp_id: 5)</li>
<li>Only emp_id 5 is included as they have the second-highest salary</li>
</ul>
</li>
<li><strong>Marketing Department</strong>:
<ul>
<li>Highest salary is 55000 (emp_id: 9)</li>
<li>Second-highest salary is 50000 (emp_id: 8)</li>
<li>Employee 8&nbsp;is included</li>
</ul>
</li>
<li><strong>HR Department</strong>:
<ul>
<li>Only has one employee</li>
<li>Not included in the result as it has fewer than 2 employees</li>
</ul>
</li>

</ul>
</div>

<!-- description:end -->

## Solutions

<!-- solution:start -->

### Solution 1: Window Function

We can use the `DENSE_RANK()` window function to rank employees in each department by salary in descending order, and then filter out the employees with a rank of $2$.

<!-- tabs:start -->

#### MySQL

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT
emp_id,
dept,
DENSE_RANK() OVER (
PARTITION BY dept
ORDER BY salary DESC
) rk
FROM Employees
)
SELECT emp_id, dept
FROM T
WHERE rk = 2
ORDER BY 1;
```

#### Pandas

```python
import pandas as pd


def find_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
employees["rk"] = employees.groupby("dept")["salary"].rank(
method="dense", ascending=False
)
second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
return second_highest.sort_values(by="emp_id")
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
9 changes: 9 additions & 0 deletions solution/3300-3399/3338.Second Highest Salary II/Solution.py
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import pandas as pd


def find_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
employees["rk"] = employees.groupby("dept")["salary"].rank(
method="dense", ascending=False
)
second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
return second_highest.sort_values(by="emp_id")
16 changes: 16 additions & 0 deletions solution/3300-3399/3338.Second Highest Salary II/Solution.sql
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# Write your MySQL query statement below
WITH
T AS (
SELECT
emp_id,
dept,
DENSE_RANK() OVER (
PARTITION BY dept
ORDER BY salary DESC
) rk
FROM Employees
)
SELECT emp_id, dept
FROM T
WHERE rk = 2
ORDER BY 1;
1 change: 1 addition & 0 deletions solution/DATABASE_README.md
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| 3308 | [寻找表现最佳的司机](/solution/3300-3399/3308.Find%20Top%20Performing%20Driver/README.md) | `数据库` | 中等 | 🔒 |
| 3322 | [英超积分榜排名 III](/solution/3300-3399/3322.Premier%20League%20Table%20Ranking%20III/README.md) | `数据库` | 中等 | 🔒 |
| 3328 | [查找每个州的城市 II](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README.md) | `数据库` | 中等 | 🔒 |
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md) | `数据库` | 中等 | 🔒 |

## 版权

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1 change: 1 addition & 0 deletions solution/DATABASE_README_EN.md
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Expand Up @@ -297,6 +297,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
| 3308 | [Find Top Performing Driver](/solution/3300-3399/3308.Find%20Top%20Performing%20Driver/README_EN.md) | `Database` | Medium | 🔒 |
| 3322 | [Premier League Table Ranking III](/solution/3300-3399/3322.Premier%20League%20Table%20Ranking%20III/README_EN.md) | `Database` | Medium | 🔒 |
| 3328 | [Find Cities in Each State II](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md) | `Database` | Medium | 🔒 |
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md) | `Database` | Medium | 🔒 |

## Copyright

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1 change: 1 addition & 0 deletions solution/README.md
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| 3335 | [字符串转换后的长度 I](/solution/3300-3399/3335.Total%20Characters%20in%20String%20After%20Transformations%20I/README.md) | | 中等 | 第 421 场周赛 |
| 3336 | [最大公约数相等的子序列数量](/solution/3300-3399/3336.Find%20the%20Number%20of%20Subsequences%20With%20Equal%20GCD/README.md) | | 困难 | 第 421 场周赛 |
| 3337 | [字符串转换后的长度 II](/solution/3300-3399/3337.Total%20Characters%20in%20String%20After%20Transformations%20II/README.md) | | 困难 | 第 421 场周赛 |
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md) | `数据库` | 中等 | 🔒 |

## 版权

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1 change: 1 addition & 0 deletions solution/README_EN.md
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Expand Up @@ -3346,6 +3346,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
| 3335 | [Total Characters in String After Transformations I](/solution/3300-3399/3335.Total%20Characters%20in%20String%20After%20Transformations%20I/README_EN.md) | | Medium | Weekly Contest 421 |
| 3336 | [Find the Number of Subsequences With Equal GCD](/solution/3300-3399/3336.Find%20the%20Number%20of%20Subsequences%20With%20Equal%20GCD/README_EN.md) | | Hard | Weekly Contest 421 |
| 3337 | [Total Characters in String After Transformations II](/solution/3300-3399/3337.Total%20Characters%20in%20String%20After%20Transformations%20II/README_EN.md) | | Hard | Weekly Contest 421 |
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md) | `Database` | Medium | 🔒 |

## Copyright

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