184. Department Highest Salary

Description

Table: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id is the primary key (column with unique values) for this table.
departmentId is a foreign key (reference columns) of the ID from the Department table.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.

 

Table: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id is the primary key (column with unique values) for this table. It is guaranteed that department name is not NULL.
Each row of this table indicates the ID of a department and its name.

 

Write a solution to find employees who have the highest salary in each of the departments.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 70000  | 1            |
| 2  | Jim   | 90000  | 1            |
| 3  | Henry | 80000  | 2            |
| 4  | Sam   | 60000  | 2            |
| 5  | Max   | 90000  | 1            |
+----+-------+--------+--------------+
Department table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Jim      | 90000  |
| Sales      | Henry    | 80000  |
| IT         | Max      | 90000  |
+------------+----------+--------+
Explanation: Max and Jim both have the highest salary in the IT department and Henry has the highest salary in the Sales department.

Solutions

Solution 1: Equi-Join + Subquery

We can use an equi-join to join the Employee table and the Department table based on Employee.departmentId = Department.id, and then use a subquery to find the highest salary for each department. Finally, we can use a WHERE clause to filter out the employees with the highest salary in each department.

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# Write your MySQL query statement below
SELECT d.name AS department, e.name AS employee, salary
FROM
    Employee AS e
    JOIN Department AS d ON e.departmentId = d.id
WHERE
    (d.id, salary) IN (
        SELECT departmentId, MAX(salary)
        FROM Employee
        GROUP BY 1
    );

Solution 2: Equi-Join + Window Function

We can use an equi-join to join the Employee table and the Department table based on Employee.departmentId = Department.id, and then use the window function rank(), which assigns a rank to each employee in each department based on their salary. Finally, we can select the rows with a rank of $1$ for each department.

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# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            d.name AS department,
            e.name AS employee,
            salary,
            RANK() OVER (
                PARTITION BY d.name
                ORDER BY salary DESC
            ) AS rk
        FROM
            Employee AS e
            JOIN Department AS d ON e.departmentId = d.id
    )
SELECT department, employee, salary
FROM T
WHERE rk = 1;