[LeetCode MySQL] 184. Department Highest Salary

184. Department Highest Salary

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

+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

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

Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, your SQL query should return the following rows (order of rows does not matter).

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
+------------+----------+--------+

Explanation:

Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

 

Solution 1 - MAX

select Department, Employee, Salary
from (select d.name Department,
             e.name Employee,
             e.salary Salary,
             max(salary) over (partition by d.id) max_salary
      from employee e inner join department d
      on e.DepartmentId = d.Id) result
where salary = max_salary

Solution 2 - RANK

select Department, Employee, Salary
from (select d.name Department,
             e.name Employee,
             Salary,
             rank() over (partition by departmentid order by salary desc) ranking
      from employee e inner join department d
      on e.DepartmentId = d.Id) result
where ranking = 1

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Reference: 184. Department Highest Salary