# leetcode\_185

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   | 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.

+----+----------+
| 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).

+------------+----------+--------+
| 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

* Caution, the number of top three employees may be larger than 3, since some employees earns the same amount of salary.
* **inner join**

```sql
# Write your MySQL query statement below
SELECT
    D.Name AS Department,
    E.Name AS Employee,
    E.Salary
FROM Employee E, Department D
WHERE E.DepartmentId = D.Id
    AND 3 > (
        SELECT COUNT(DISTINCT E2.Salary)
        FROM Employee E2
        WHERE E2.DepartmentId = E.DepartmentId
            AND E2.Salary > E.Salary
    )
ORDER BY E.DepartmentId, E.Salary DESC
```

1. **having**

```sql
```

1. **dense\_rank**

```sql
# Write your MySQL query statement below

SELECT Department, Employee, Salary
FROM (
    SELECT 
        D.Name AS Department,
        E.Name AS Employee,
        Salary,
        DENSE_RANK() OVER(PARTITION BY E.DepartmentId ORDER BY Salary DESC) AS Rank
    FROM Employee E, Department D
    WHERE E.DepartmentId = D.Id)
    AS S
WHERE S.Rank <= 3;
```