How do you solve Number of Students Doing Homework at a Given Time on LeetCode?

Number of Students Doing Homework at a Given Time (LeetCode #1450) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding intervals is crucial; queryTime must be checked against both start and end times.

What is the brute force solution for Number of Students Doing Homework at a Given Time?

The brute force approach for Number of Students Doing Homework at a Given Time is Brute Force, with O(n) time complexity and O(1) space complexity. We can simply check each student's start and end time to see if the queryTime falls within that range. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Students Doing Homework at a Given Time?

Number of Students Doing Homework at a Given Time uses the following concepts: Array. The recommended patterns to study are: Array, Interval Checking.

What are the interview tips for Number of Students Doing Homework at a Given Time?

Always clarify the problem and constraints before jumping into coding. Think about edge cases and test your solution against them. Explain your thought process while coding; it shows your understanding and helps interviewers follow your logic.

What are common mistakes when solving Number of Students Doing Homework at a Given Time?

Failing to include the boundaries (inclusive checks) when comparing queryTime with startTime and endTime. Not considering edge cases, such as when only one student is present.

#1450

Number of Students Doing Homework at a Given Time

Easy

ArrayArrayInterval Checking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The brute force approach is already efficient for the given constraints, but we can ensure clarity and maintainability by using a more structured approach with list comprehensions or functional programming.

⚙️

Algorithm

2 steps

1Step 1: Use a list comprehension or filter to count students whose queryTime falls within their start and end times.
2Step 2: Return the length of the resulting list.

solution.py2 lines

1def busyStudents(startTime, endTime, queryTime):
2    return sum(1 for s, e in zip(startTime, endTime) if s <= queryTime <= e)

ℹ

Complexity note: The time complexity remains O(n) since we still iterate through the list of students once. The space complexity is O(1) as we do not use any additional data structures that grow with input size.

1Understanding intervals is crucial; queryTime must be checked against both start and end times.
2Using list comprehensions or functional programming can lead to cleaner and more maintainable code.

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