-
Notifications
You must be signed in to change notification settings - Fork 0
/
Department Highest Salary.sql
47 lines (40 loc) · 1.55 KB
/
Department Highest Salary.sql
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
/*
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 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
The Department table holds all departments of the company.
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
Link: https://leetcode.com/problems/department-highest-salary/
Example: Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables,
Max has the highest salary in the IT department and Henry has the highest salary in the Sales department.
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
Solution: None
Source: https://leetcode.com/discuss/49168/easy-solution-no-joins-group-by-is-enough-916ms
*/
# Write your MySQL query statement below
select
d.Name, e.Name, e.Salary
from
Department d,
Employee e,
(select MAX(Salary) as Salary, DepartmentId as DepartmentId from Employee GROUP BY DepartmentId) h
where
e.Salary = h.Salary and
e.DepartmentId = h.DepartmentId and
e.DepartmentId = d.Id;