39-00184-department-highest-salary.sql

184. Department Highest Salary

Table: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id is the primary key (column with unique values) for this table.
departmentId is a foreign key (reference columns) 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 with unique values) for this table. It is guaranteed that department name is not NULL.
Each row of this table indicates the ID of a department and its name.

 

Write a solution to find employees who have the highest salary in each of the departments.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
Explanation: Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

-- SQL Schema

Create table If Not Exists Employee (id int, name varchar(255), salary int, departmentId int)
Create table If Not Exists Department (id int, name varchar(255))
Truncate table Employee
insert into Employee (id, name, salary, departmentId) values ('1', 'Joe', '70000', '1')
insert into Employee (id, name, salary, departmentId) values ('2', 'Jim', '90000', '1')
insert into Employee (id, name, salary, departmentId) values ('3', 'Henry', '80000', '2')
insert into Employee (id, name, salary, departmentId) values ('4', 'Sam', '60000', '2')
insert into Employee (id, name, salary, departmentId) values ('5', 'Max', '90000', '1')
Truncate table Department
insert into Department (id, name) values ('1', 'IT')
insert into Department (id, name) values ('2', 'Sales')

-- Solution
----------------------------------------------------------------------------------------------------------------
--Oracle & MySQL
----------------------------------------------------------------------------------------------------------------
 
-- using subquery

select d.name as Department, e.name as Employee, e.salary
from Department d join Employee e
on d.id = e.departmentId
where (e.departmentId, e.salary) in 
    (select departmentId, max(salary)
    from Employee
    group by departmentId)

-----------------------------------------------------------------------------------------------------------------------------------------------------------

-- using window function

with CTE as
(select departmentId, id, name, salary,
    dense_rank() over(partition by departmentId order by salary desc) as rnk
from Employee)

select d.name as Department, e.name as Employee, e.salary
from Department d join CTE e
on d.id = e.departmentId
where e.rnk = 1


-- amazon- 2
-- microsoft- 3
-- apple- 2
-- facebook- 2
-- google- 2