LeetCode181. Employees Earning More Than Their Managers

Table: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| salary      | int     |
| managerId   | int     |
+-------------+---------+
id is the primary key column for this table.
Each row of this table indicates the ID of an employee, their name, salary, and the ID of their manager.

 

Write an SQL query to find the employees who earn more than their managers.

Return the result table in any order.

The query result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+-------+--------+-----------+
| id | name  | salary | managerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | Null      |
| 4  | Max   | 90000  | Null      |
+----+-------+--------+-----------+
Output: 
+----------+
| Employee |
+----------+
| Joe      |
+----------+
Explanation: Joe is the only employee who earns more than his manager.

Create table If Not Exists Employee (id int, name varchar(255), salary int, managerId int)

Truncate table Employee

insert into Employee (id, name, salary, managerId) values ('1', 'Joe', '70000', '3')

insert into Employee (id, name, salary, managerId) values ('2', 'Henry', '80000', '4')

insert into Employee (id, name, salary, managerId) values ('3', 'Sam', '60000', 'None')

insert into Employee (id, name, salary, managerId) values ('4', 'Max', '90000', 'None')

select E.name "Employee"

from Employee E join Employee M

ON E.managerId = M.id

where E.salary > M.salary