Employee Bonus - Practice Coding Problems

Employee Bonus - Problem

Database Easy

Given two tables: Employee and Bonus, write a SQL solution to report the name and bonus amount of each employee who satisfies either of the following conditions:

The employee has a bonus less than 1000
The employee did not get any bonus

Return the result table in any order.

Employee Table Schema:

The Employee table contains employee information including their ID, name, supervisor, and salary.

Bonus Table Schema:

The Bonus table contains the bonus amount for some employees. Not all employees may have an entry in this table.

Table Schema

Employee

Column Name	Type	Description
`empId` PK	int	Unique employee ID
`name`	varchar	Employee name
`supervisor`	int	Supervisor employee ID
`salary`	int	Employee salary

Primary Key: empId

Bonus

Column Name	Type	Description
`empId` PK	int	Employee ID (foreign key)
`bonus`	int	Bonus amount

Primary Key: empId

Input & Output

Example 1 — Mixed Bonus Scenarios

Input Tables:

Employee

empId	name	supervisor	salary
3	Brad		4000
1	John	3	1000
2	Dan	3	2000
4	Thomas	3	4000

Bonus

empId	bonus
2	500
4	2000

Output:

name	bonus
Brad
John
Dan	500

💡 Note:

Brad and John have no bonus records (NULL), so they qualify. Dan has bonus 500 < 1000, so he qualifies. Thomas has bonus 2000 ≥ 1000, so he's excluded.

Example 2 — All Employees No Bonus

Input Tables:

Employee

empId	name	supervisor	salary
1	Alice		3000
2	Bob	1	2500

Bonus

empId	bonus

Output:

name	bonus
Alice
Bob

💡 Note:

No employees have bonus records, so all employees qualify with NULL bonus values.

Constraints

1 ≤ empId ≤ 500
name consists of lowercase English letters
1 ≤ salary ≤ 10000
1 ≤ bonus ≤ 10000

Visualization

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Understanding the Visualization

Input Tables

Employee and Bonus tables

LEFT JOIN

Include all employees

Filter Results

bonus < 1000 OR NULL

Key Takeaway

🎯 Key Insight: Use LEFT JOIN when you need to include records that may not exist in the second table

Asked in

F Facebook 15 A Amazon 12 L LinkedIn 8

Use LEFT JOIN to combine Employee and Bonus tables, ensuring all employees are included. Filter with WHERE bonus < 1000 OR bonus IS NULL to find qualifying employees.

Table Schema

Employee

Column Name	Type	Description
`empId` PK	int	Unique employee ID
`name`	varchar	Employee name
`supervisor`	int	Supervisor employee ID
`salary`	int	Employee salary

Primary Key: empId

Bonus

Column Name	Type	Description
`empId` PK	int	Employee ID (foreign key)
`bonus`	int	Bonus amount

Primary Key: empId

Common Approaches

Approach	Time	Space	Notes
✓ Left Join with Conditional Filtering	O(n + m)	O(n)	Use LEFT JOIN to include employees without bonuses and filter by conditions

Left Join with Conditional Filtering — Algorithm Steps

Step 1: LEFT JOIN Employee with Bonus on empId
Step 2: Filter WHERE bonus < 1000 OR bonus IS NULL

Visualization

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Step-by-Step Walkthrough

Left Join

Include all employees

Filter

bonus < 1000 OR NULL

Select

name and bonus

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Linear scan of both tables with hash join

✓ Linear Growth

Space Complexity

O(n)

Space for result set

⚡ Linearithmic Space

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Smart Actions

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Table Schema

Input & Output

Constraints

Visualization

Related Problems

Table Schema

Common Approaches

Left Join with Conditional Filtering — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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