How do you solve Employee Bonus on LeetCode?

Employee Bonus (LeetCode #577) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using LEFT JOIN ensures we capture all employees, even those without bonuses.

What is the time complexity of Employee Bonus?

The optimal solution for Employee Bonus runs in O(n) time and uses O(n) space. This complexity is linear because we are processing each employee and their corresponding bonus in a single pass.

What is the brute force solution for Employee Bonus?

The brute force approach for Employee Bonus is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each employee's bonus against the conditions specified. If they don't have a bonus or their bonus is less than 1000, we will include them in the results. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Employee Bonus?

Employee Bonus uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Employee Bonus?

Always clarify the requirements before jumping into coding. Think about edge cases, such as employees without bonuses. Practice writing SQL queries on paper to get comfortable with syntax.

What are common mistakes when solving Employee Bonus?

Forgetting to use LEFT JOIN, which can lead to missing employees without bonuses. Not considering null as a valid condition for the bonus.

#577

Employee Bonus

Easy

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can efficiently retrieve the required information using a LEFT JOIN, which allows us to get all employees and their bonuses in one go. We can then filter the results based on the conditions.

⚙️

Algorithm

3 steps

1Step 1: Perform a LEFT JOIN between Employee and Bonus tables on empId.
2Step 2: Use COALESCE to handle null bonuses, treating them as null.
3Step 3: Filter results where the bonus is either less than 1000 or null.

solution.py5 lines

1# Full working Python code
2SELECT e.name, COALESCE(b.bonus, NULL) AS bonus
3FROM Employee e
4LEFT JOIN Bonus b ON e.empId = b.empId
5WHERE b.bonus < 1000 OR b.bonus IS NULL;

ℹ

Complexity note: This complexity is linear because we are processing each employee and their corresponding bonus in a single pass.

1Using LEFT JOIN ensures we capture all employees, even those without bonuses.
2COALESCE is useful for handling null values effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.