185. Department Top Three Salaries (Hard)

Posted 2020-09-19 sixu

Source: https://leetcode.com/problems/department-top-three-salaries/#/description

Description:

The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.

+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+

Write a SQL query to find employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows.

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+

 

Solution:

select t2.name  as Department, t1.Employee , t1.salary from 
(
    select e1.id, e1.name as employee, e1.salary, e1.departmentid from Employee  e1
    inner join  Employee e2
    on e1.departmentid = e2.departmentid
    and e1.salary<=e2.salary
    group by e1.id
    having count(distinct(e2.salary))<=3
)
t1
inner join
(select * from Department )t2
on t1.DepartmentId  = t2.id