@@ -1,3 +1,74 @@
+-- Problem:
+
+/*
+Table: Employee
+
++-------------+---------+
+| Column Name | Type |
++-------------+---------+
+| empId | int |
+| name | varchar |
+| supervisor | int |
+| salary | int |
++-------------+---------+
+empId is the column with unique values 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 column of unique values for this table.
+empId is a foreign key (reference column) to empId from the Employee table.
+Each row of this table contains the id of an employee and their respective bonus.
+
+
+Write a solution to report the name and bonus amount of each employee with a bonus less than 1000.
+
+Return the result table in any order.
+
+The 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 e.name , b.bonus
 FROM Employee e
 LEFT JOIN Bonus b ON e.empId=b.empId