How do you solve Find Overbooked Employees on LeetCode?

Find Overbooked Employees (LeetCode #3611) 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: Meeting hours can be aggregated by week.

What is the time complexity of Find Overbooked Employees?

The optimal solution for Find Overbooked Employees runs in O(n) time and uses O(n) space. Single pass aggregation leads to linear time complexity, with space used for storing intermediate results.

What is the brute force solution for Find Overbooked Employees?

The brute force approach for Find Overbooked Employees is Brute Force, with O(n²) time complexity and O(1) space complexity. Calculate meeting hours for each employee for every week, then compare to total working hours. This involves checking each meeting against all employees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Overbooked Employees?

Find Overbooked Employees uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Overbooked Employees?

Clarify assumptions about work hours. Discuss edge cases like holidays. Explain your thought process clearly.

What are common mistakes when solving Find Overbooked Employees?

Not grouping meetings by week. Confusing total hours with weekly hours.

#3611

Find Overbooked Employees

Medium

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)

Use a single pass to aggregate meeting hours by week for each employee, then filter those exceeding 20 hours. This reduces redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Group meetings by employee and week, summing durations.
2Step 2: Filter employees whose total meeting hours exceed 20 hours.
3Step 3: Return the list of overbooked employees.

solution.py1 lines

1SELECT employee_id, employee_name FROM employees e JOIN (SELECT employee_id, YEAR(meeting_date) AS year, WEEK(meeting_date) AS week, SUM(duration_hours) AS total_hours FROM meetings GROUP BY employee_id, year, week) m ON e.employee_id = m.employee_id WHERE total_hours > 20

ℹ

Complexity note: Single pass aggregation leads to linear time complexity, with space used for storing intermediate results.

1Meeting hours can be aggregated by week.
2Filtering after aggregation is efficient.

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