---
# System prepended metadata

title: Leet code sql 185. Department Top Three Salaries
tags: [LeetCode, ' coding']

---

---
tags: LeetCode, coding
---

# Leet code sql 185. Department Top Three Salaries
## description
```sql
# sql scheme
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', '85000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('2', 'Henry', '80000', '2')
insert into Employee (Id, Name, Salary, DepartmentId) values ('3', 'Sam', '60000', '2')
insert into Employee (Id, Name, Salary, DepartmentId) values ('4', 'Max', '90000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('5', 'Janet', '69000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('6', 'Randy', '85000', '1')
insert into Employee (Id, Name, Salary, DepartmentId) values ('7', 'Will', '70000', '1')
Truncate table Department
insert into Department (Id, Name) values ('1', 'IT')
insert into Department (Id, Name) values ('2', 'Sales')
```

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

```sql
+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+
```

The Department table holds all departments of the company.

```sql
+----+----------+
| 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 (order of rows does not matter).

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

Explanation:

In IT department, Max earns the highest salary, both Randy and Joe earn the second highest salary, and Will earns the third highest salary. There are only two employees in the Sales department, Henry earns the highest salary while Sam earns the second highest salary.

## solutions
1. 官方提供的解答

```sql
SELECT e1.Name as Name, e1.Salary as Salary, d.Name as Department
FROM `Employee` e1 JOIN `Department` d on e1.DepartmentId = d.Id
WHERE 3 > (
    SELECT COUNT(DISTINCT(e2.Salary))
    FROM `Employee` e2
    WHERE e1.DepartmentId = e2.DepartmentId AND e1.Salary < e2.Salary
);

# output
+-------+--------+------------+
| Name  | Salary | Department |
+-------+--------+------------+
| Joe   |  85000 | IT         |
| Henry |  80000 | Sales      |
| Sam   |  60000 | Sales      |
| Max   |  90000 | IT         |
| Randy |  85000 | IT         |
| Will  |  70000 | IT         |
+-------+--------+------------+
```

2. 自己的解答，有另外做出分組排名欄位，使報表更一目了然
>    1. 先做出分群排名表(這邊group by id是為了取每個薪水數字)

```sql
SELECT e1.id, e1.Name as Name, e1.Salary as Salary, d.Name as Department, COUNT(DISTINCT(e2.Salary)) as Group_rank
FROM `Employee` e1 JOIN `Department` d on e1.DepartmentId = d.Id, `Employee` e2
WHERE e1.DepartmentId = e2.DepartmentId AND e1.Salary <= e2.Salary
GROUP BY e1.Id;

# output
+------+-------+--------+------------+------------+
| id   | Name  | Salary | Department | Group_rank |
+------+-------+--------+------------+------------+
|    1 | Joe   |  85000 | IT         |          2 |
|    2 | Henry |  80000 | Sales      |          1 |
|    3 | Sam   |  60000 | Sales      |          2 |
|    4 | Max   |  90000 | IT         |          1 |
|    5 | Janet |  69000 | IT         |          4 |
|    6 | Randy |  85000 | IT         |          2 |
|    7 | Will  |  70000 | IT         |          3 |
+------+-------+--------+------------+------------+
```

>    2. 再進一步JOIN Department 

```sql
SELECT Name, Salary, Department, Group_rank
FROM (
    SELECT e1.id, e1.Name as Name, e1.Salary as Salary, d.Name as Department, COUNT(DISTINCT(e2.Salary)) as Group_rank
    FROM `Employee` e1 JOIN `Department` d on e1.DepartmentId = d.Id, `Employee` e2
    WHERE e1.DepartmentId = e2.DepartmentId AND e1.Salary <= e2.Salary
    GROUP BY e1.Id
) temp WHERE Group_rank <= 3;

# output
+-------+--------+------------+------------+
| Name  | Salary | Department | Group_rank |
+-------+--------+------------+------------+
| Joe   |  85000 | IT         |          2 |
| Henry |  80000 | Sales      |          1 |
| Sam   |  60000 | Sales      |          2 |
| Max   |  90000 | IT         |          1 |
| Randy |  85000 | IT         |          2 |
| Will  |  70000 | IT         |          3 |
+-------+--------+------------+------------+
```
