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feat: add solutions to lc problem: No.3338 (doocs#3684)
No.3338.Second Highest Salary II
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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/3300-3399/3338.Second%20Highest%20Salary%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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# [3338. Second Highest Salary II 🔒](https://leetcode.cn/problems/second-highest-salary-ii)
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[English Version](/solution/3300-3399/3338.Second%20Highest%20Salary%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>employees</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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| emp_id | int |
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| salary | int |
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| dept | varchar |
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+------------------+---------+
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emp_id is the unique key for this table.
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Each row of this table contains information about an employee including their ID, name, manager, salary, department, start date, and building assignment.
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</pre>
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<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>
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<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>
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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>employees table:</p>
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<pre class="example-io">
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+--------+--------+-----------+
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| emp_id | salary | dept |
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+--------+--------+-----------+
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| 1 | 70000 | Sales |
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| 2 | 80000 | Sales |
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| 3 | 80000 | Sales |
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| 4 | 90000 | Sales |
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| 5 | 55000 | IT |
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| 6 | 65000 | IT |
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| 7 | 65000 | IT |
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| 8 | 50000 | Marketing |
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| 9 | 55000 | Marketing |
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| 10 | 55000 | HR |
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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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| emp_id | dept |
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+--------+-----------+
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| 2 | Sales |
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| 3 | Sales |
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| 5 | IT |
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| 8 | Marketing |
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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>Sales Department</strong>:
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<ul>
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<li>Highest salary is 90000 (emp_id: 4)</li>
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<li>Second-highest salary is 80000 (emp_id: 2, 3)</li>
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<li>Both employees with salary 80000 are included</li>
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</ul>
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</li>
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<li><strong>IT Department</strong>:
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<ul>
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<li>Highest salary is 65000 (emp_id: 6, 7)</li>
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<li>Second-highest salary is 55000 (emp_id: 5)</li>
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<li>Only emp_id 5 is included as they have the second-highest salary</li>
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</ul>
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</li>
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<li><strong>Marketing Department</strong>:
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<ul>
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<li>Highest salary is 55000 (emp_id: 9)</li>
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<li>Second-highest salary is 50000 (emp_id: 8)</li>
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<li>Employee 8&nbsp;is included</li>
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</ul>
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</li>
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<li><strong>HR Department</strong>:
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<ul>
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<li>Only has one employee</li>
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<li>Not included in the result as it has fewer than 2 employees</li>
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</ul>
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</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:窗口函数
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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ドル$ 的员工即可。
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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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WITH
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T AS (
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SELECT
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emp_id,
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dept,
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DENSE_RANK() OVER (
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PARTITION BY dept
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ORDER BY salary DESC
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) rk
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FROM Employees
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)
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SELECT emp_id, dept
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FROM T
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WHERE rk = 2
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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_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
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employees["rk"] = employees.groupby("dept")["salary"].rank(
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method="dense", ascending=False
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)
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second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
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return second_highest.sort_values(by="emp_id")
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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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---
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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/3300-3399/3338.Second%20Highest%20Salary%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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# [3338. Second Highest Salary II 🔒](https://leetcode.com/problems/second-highest-salary-ii)
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[中文文档](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code>employees</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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| emp_id | int |
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| salary | int |
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| dept | varchar |
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+------------------+---------+
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emp_id is the unique key for this table.
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Each row of this table contains information about an employee including their ID, name, manager, salary, department, start date, and building assignment.
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</pre>
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<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>
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<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>
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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>employees table:</p>
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<pre class="example-io">
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+--------+--------+-----------+
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| emp_id | salary | dept |
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+--------+--------+-----------+
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| 1 | 70000 | Sales |
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| 2 | 80000 | Sales |
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| 3 | 80000 | Sales |
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| 4 | 90000 | Sales |
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| 5 | 55000 | IT |
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| 6 | 65000 | IT |
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| 7 | 65000 | IT |
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| 8 | 50000 | Marketing |
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| 9 | 55000 | Marketing |
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| 10 | 55000 | HR |
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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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| emp_id | dept |
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+--------+-----------+
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| 2 | Sales |
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| 3 | Sales |
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| 5 | IT |
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| 8 | Marketing |
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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>Sales Department</strong>:
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<ul>
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<li>Highest salary is 90000 (emp_id: 4)</li>
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<li>Second-highest salary is 80000 (emp_id: 2, 3)</li>
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<li>Both employees with salary 80000 are included</li>
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</ul>
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</li>
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<li><strong>IT Department</strong>:
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<ul>
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<li>Highest salary is 65000 (emp_id: 6, 7)</li>
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<li>Second-highest salary is 55000 (emp_id: 5)</li>
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<li>Only emp_id 5 is included as they have the second-highest salary</li>
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</ul>
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</li>
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<li><strong>Marketing Department</strong>:
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<ul>
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<li>Highest salary is 55000 (emp_id: 9)</li>
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<li>Second-highest salary is 50000 (emp_id: 8)</li>
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<li>Employee 8&nbsp;is included</li>
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</ul>
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</li>
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<li><strong>HR Department</strong>:
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<ul>
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<li>Only has one employee</li>
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<li>Not included in the result as it has fewer than 2 employees</li>
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</ul>
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</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: Window Function
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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ドル$.
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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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WITH
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T AS (
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SELECT
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emp_id,
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dept,
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DENSE_RANK() OVER (
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PARTITION BY dept
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ORDER BY salary DESC
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) rk
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FROM Employees
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)
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SELECT emp_id, dept
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FROM T
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WHERE rk = 2
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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_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
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employees["rk"] = employees.groupby("dept")["salary"].rank(
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method="dense", ascending=False
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)
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second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
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return second_highest.sort_values(by="emp_id")
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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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import pandas as pd
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def find_second_highest_salary(employees: pd.DataFrame) -> pd.DataFrame:
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employees["rk"] = employees.groupby("dept")["salary"].rank(
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method="dense", ascending=False
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)
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second_highest = employees[employees["rk"] == 2][["emp_id", "dept"]]
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return second_highest.sort_values(by="emp_id")
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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emp_id,
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dept,
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DENSE_RANK() OVER (
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PARTITION BY dept
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ORDER BY salary DESC
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) rk
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FROM Employees
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)
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SELECT emp_id, dept
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FROM T
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WHERE rk = 2
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ORDER BY 1;

‎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 | 🔒 |
300300
| 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 | 🔒 |
301301
| 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 | 🔒 |
302+
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md) | `数据库` | 中等 | 🔒 |
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## 版权
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‎solution/DATABASE_README_EN.md‎

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| 3308 | [Find Top Performing Driver](/solution/3300-3399/3308.Find%20Top%20Performing%20Driver/README_EN.md) | `Database` | Medium | 🔒 |
298298
| 3322 | [Premier League Table Ranking III](/solution/3300-3399/3322.Premier%20League%20Table%20Ranking%20III/README_EN.md) | `Database` | Medium | 🔒 |
299299
| 3328 | [Find Cities in Each State II](/solution/3300-3399/3328.Find%20Cities%20in%20Each%20State%20II/README_EN.md) | `Database` | Medium | 🔒 |
300+
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md) | `Database` | Medium | 🔒 |
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## Copyright
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‎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 |
33493349
| 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 |
33503350
| 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 |
3351+
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README.md) | `数据库` | 中等 | 🔒 |
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## 版权
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‎solution/README_EN.md‎

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| 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 |
33473347
| 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 |
33483348
| 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 |
3349+
| 3338 | [Second Highest Salary II](/solution/3300-3399/3338.Second%20Highest%20Salary%20II/README_EN.md) | `Database` | Medium | 🔒 |
33493350

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## Copyright
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