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185. Department Top Three Salaries (Hard)
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
185. Department Top Three Salaries (Hard)
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