Find Employees Who Earn More Than Their Department Average

SQL coding challenge · Difficulty: medium · +100 XP

Problem

A compensation analyst wants to find employees who earn above their own department's average — high performers who are strong candidates for promotion.

Tables

Table: departments

| department_id | department_name |
| --- | --- |
| 1 | Engineering |
| 2 | Marketing |
| 3 | HR |
| 4 | Finance |

Table: employees

| employee_id | first_name | department_id | salary | hire_date | manager_id |
| --- | --- | --- | --- | --- | --- |
| 1 | Alice | 1 | 90000 | 2021-03-10 | NULL |
| 2 | Bob | 1 | 75000 | 2021-03-20 | 1 |
| 3 | Charlie | 1 | 80000 | 2021-03-25 | 1 |
| 4 | Victor | 1 | 65000 | 2022-01-15 | 1 |
| 5 | Diana | 2 | 70000 | 2022-06-05 | 1 |
| 6 | Eve | 2 | 65000 | 2021-11-15 | 5 |
| 7 | Frank | 3 | 60000 | 2023-02-28 | 1 |

Expected Output

| employee_id | first_name | salary | department_id |
| --- | --- | --- | --- |
| 1 | Alice | 90000 | 1 |
| 3 | Charlie | 80000 | 1 |
| 5 | Diana | 70000 | 2 |

• Return: employee_id, first_name, salary, department_id
• Only employees earning above their own department's average
• Sort by salary descending
• Hint: Use a correlated subquery: WHERE salary > (SELECT AVG(salary) FROM employees e2 WHERE e2.department_id = e1.department_id)