SQL : Department Top Three Salaries

 Hello All,

We are back with another SQL problem to gain understanding about SQL. This blog-post focuses on implementation of sub-queries in FROM clause. The problem below is sourced from Leetcode. You can visit the link: Department Top Three Salaries - LeetCode

To know more about SQL sub-query in FROM clause , I highly recommend this blog-post.

SQL | Sub queries in From Clause - GeeksforGeeks

Give a try to solve below problem. Solution is provided as usual. Feel free to comment if any queries and also suggest alternate approach to solve this problem. Have a Happy Learning !  

SQL Schema

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.

 

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:

SELECT Distinct Department, Employee, Salary
from(
    SELECT dep.name as Department, emp.name as Employee, emp.Salary as Salary,

    DENSE_RANK() OVER (PARTITION BY dep.name order by emp.salary desc) as r
    FROM Department as dep
    Join Employee as emp
    on dep.id = emp.departmentId
) as temp
WHERE r <= 3

Explanation :

In the solution, DENSE_RANK, a SQL window function has been used. Previously we had solved a easy conceptual problem using window function DENSE_RANK(). You can visit following link to know more about difference between RANK() and DENSE_RANK().

2. Window functions in SQL - GeeksforGeeks

As you see in problem, name is ambiguous as it refers to both Department name and Employee name. So, emp.name and dep.name refers to them respectively. 

Further DENSE_RANK is used with OVER clause, which partitioned by PARTITION BY clause. Here, what PARTITION clause do is creating subset of rows based on the department name. And then, salaries are arranged in descending order as:

DENSE_RANK() OVER (PARTITION BY dep.name order by emp.salary desc) 

We have used INNER JOIN which can denoted simply as JOIN to connect these 2 tables over the columns dep.id and emp.departmentId .

Finally WHERE clause is used to fetch top 3 salaries in each department.

Happy Learning !

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