Employee Bonus

LeetCode 577 | Difficulty: Easy

Easy

Problem Description

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 who satisfies either of the following:

- The employee has a bonus **less than** `1000`.

- The employee did not get any bonus.

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

Topics: Database

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Approach

Direct Approach

This problem can typically be solved with straightforward iteration or simple data structure usage. Focus on correctness first, then optimize.

When to use

Basic problems that test fundamental programming skills.

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Solutions

Solution 1: MySQL (Best: 172 ms)

Metric	Value
Runtime	172 ms
Memory	N/A
Date	2018-05-08

Solution

# Write your MySQL query statement below
select A.name , B.bonus from Employee A left outer join Bonus B 
on A.empid = B.empid
WHERE B.bonus <  1000 OR B.bonus is null

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• LeetCode provides 2 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: If the EmpId in table Employee has no match in table Bonus, we consider that the corresponding bonus is null and null is smaller than 1000.

Hint 2: Inner join is the default join, we can solve the mismatching problem by using outer join.