Problem
Table: Employee
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| id | int |
| name | varchar |
| salary | int |
| departmentId | int |
+--------------+---------+
id is the primary key column for this table.
departmentId is a foreign key of the ID from the Department
table.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.
Table: Department
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
+-------------+---------+
id is the primary key column for this table. Each row of this table indicates the ID of a department and its name.
A company’s executives are interested in seeing who earns the most money in each of the company’s departments. A high earner in a department is an employee who has a salary in the top three unique salaries for that department.
Write an SQL query to find the employees who are high earners in each of the departments.
Return the result table in any order.
The query result format is in the following example.
Examples
Example 1:
Input: Employee table:
+----+-------+--------+--------------+
| id | name | salary | departmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+
Department table:
+----+-------+
| id | name |
+----+-------+
| 1 | IT |
| 2 | Sales |
+----+-------+
Output:
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Joe | 85000 |
| IT | Randy | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
Explanation: In the IT department:
- Max earns the highest unique salary
- Both Randy and Joe earn the second-highest unique salary
- Will earns the third-highest unique salary
In the Sales department:
- Henry earns the highest salary
- Sam earns the second-highest salary
- There is no third-highest salary as there are only two employees
Solution
Method 1 - Using Common Table Expressions
Code
SQL
WITH RankedSalaries AS (
SELECT
e.departmentId,
e.salary,
DENSE_RANK() OVER (PARTITION BY e.departmentId ORDER BY e.salary DESC) as salaryRank
FROM
Employee e
),
TopSalaries AS (
SELECT
rs.departmentId,
rs.salary
FROM
RankedSalaries rs
WHERE
rs.salaryRank <= 3
)
SELECT DISTINCT
e.id,
e.name,
e.salary,
e.departmentId
FROM
Employee e
JOIN
TopSalaries ts
ON
e.departmentId = ts.departmentId
AND e.salary = ts.salary
ORDER BY
e.departmentId, e.salary DESC, e.id;
Complexity
- ⏰ Time complexity
DENSE_RANK
operation: The window function has a time complexity ofO(n log n)
due to sorting within each partition (department).- Join operations: The joins are
O(n log n)
in case of properly indexed tables; otherwise, in the worst case, it could beO(n^2)
.
- 🧺 Space complexity:
O(n)
- The space complexity for storing intermediate results is
O(n)
.
- The space complexity for storing intermediate results is