Leetcode(Easy) - Employee Bonus

SQL Schema

Create table If Not Exists Employee (empId int, name varchar(255), supervisor int, salary int)
Create table If Not Exists Bonus (empId int, bonus int)
Truncate table Employee
insert into Employee (empId, name, supervisor, salary) values ('3', 'Brad', 'None', '4000')
insert into Employee (empId, name, supervisor, salary) values ('1', 'John', '3', '1000')
insert into Employee (empId, name, supervisor, salary) values ('2', 'Dan', '3', '2000')
insert into Employee (empId, name, supervisor, salary) values ('4', 'Thomas', '3', '4000')
Truncate table Bonus
insert into Bonus (empId, bonus) values ('2', '500')
insert into Bonus (empId, bonus) values ('4', '2000')

Table: Employee

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

Table: Bonus

+-------------+------+
| Column Name | Type |
+-------------+------+
| empId       | int  |
| bonus       | int  |
+-------------+------+
empId is the primary key column for this table.
empId is a foreign key to empId from the Employee table.
Each row of this table contains the id of an employee and their respective bonus.

Question

Write an SQL query to report the name and bonus amount of each employee with a bonus less than 1000.

Return the result table in any order.

The query result format is in the following example.

<Example 1>

Input:
Employee table:
+-------+--------+------------+--------+
| empId | name   | supervisor | salary |
+-------+--------+------------+--------+
| 3     | Brad   | null       | 4000   |
| 1     | John   | 3          | 1000   |
| 2     | Dan    | 3          | 2000   |
| 4     | Thomas | 3          | 4000   |
+-------+--------+------------+--------+
Bonus table:
+-------+-------+
| empId | bonus |
+-------+-------+
| 2     | 500   |
| 4     | 2000  |
+-------+-------+
Output:
+------+-------+
| name | bonus |
+------+-------+
| Brad | null  |
| John | null  |
| Dan  | 500   |
+------+-------+

Solution

select name, bonus
from Employee e
left join Bonus b
using(empId)
where b.bonus < 1000 or bonus is null