Is Employee Bonus easy or hard?

Employee Bonus is classified as an Easy SQL problem with a high acceptance rate around 76%. The main challenge is remembering that employees without bonus rows produce NULL values after a LEFT JOIN and must be explicitly included in the filter.

Employee Bonus Python/Java solution

The problem is designed for SQL rather than Python or Java because it operates directly on database tables. The solution uses SQL queries such as LEFT JOIN or subqueries to combine employee records with bonus values and filter results.

How to solve Employee Bonus in O(n)?

Use a LEFT JOIN between Employee and Bonus on employee ID. After joining, apply the condition `bonus < 1000 OR bonus IS NULL`. With proper indexing on the join key, the database effectively scans the tables once, giving linear complexity relative to the dataset size.

What is the best approach for Employee Bonus?

The most common solution uses a SQL LEFT JOIN between the Employee and Bonus tables. After joining, filter rows where bonus < 1000 or bonus IS NULL. This approach is simple, readable, and runs in O(n + m) time because the database processes both tables once during the join.

Is Employee Bonus asked at Google/Amazon/Meta?

Employee Bonus represents a typical SQL interview question used by many companies including large tech firms. While the exact problem may vary, the underlying concepts—LEFT JOIN, NULL handling, and relational filtering—frequently appear in database interview rounds.

What data structure is used in Employee Bonus?

This problem primarily tests relational database concepts rather than traditional data structures. The key mechanism is a SQL JOIN operation between two tables along with NULL handling using conditions like IS NULL or COALESCE.

What is the time complexity of Employee Bonus?

The typical SQL solution runs in O(n + m) time, where n is the number of employees and m is the number of bonus records. The database performs a join or lookup across both tables. Space complexity is O(1) because the query does not require additional data structures beyond the result set.

Description Hints 2 Solution Video 4:57

577. Employee Bonus

Q: What is the best approach for Employee Bonus?

The most common solution uses a SQL LEFT JOIN between the Employee and Bonus tables. After joining, filter rows where bonus < 1000 or bonus IS NULL. This approach is simple, readable, and runs in O(n + m) time because the database processes both tables once during the join.

Q: Is Employee Bonus asked at Google/Amazon/Meta?

Employee Bonus represents a typical SQL interview question used by many companies including large tech firms. While the exact problem may vary, the underlying concepts—LEFT JOIN, NULL handling, and relational filtering—frequently appear in database interview rounds.

Q: What data structure is used in Employee Bonus?

This problem primarily tests relational database concepts rather than traditional data structures. The key mechanism is a SQL JOIN operation between two tables along with NULL handling using conditions like IS NULL or COALESCE.

Q: What is the time complexity of Employee Bonus?

The typical SQL solution runs in O(n + m) time, where n is the number of employees and m is the number of bonus records. The database performs a join or lookup across both tables. Space complexity is O(1) because the query does not require additional data structures beyond the result set.

Easy78.3% acceptance7 companies

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