Number of Students Doing Homework at a Given Time Python/Java solution

In Python or Java, iterate over the arrays and check if startTime[i] <= queryTime and queryTime <= endTime[i]. Increment a counter for each matching interval and return the result. The implementation is short, typically under 10 lines, and runs in O(n) time.

How to solve Number of Students Doing Homework at a Given Time in O(n)?

Iterate through the startTime and endTime arrays simultaneously. For each student, check if the query time lies between the start and end values inclusive. If it does, increment a counter. After scanning all students, the counter represents how many are doing homework at that moment.

What is the best approach for Number of Students Doing Homework at a Given Time?

The most practical approach is a single pass through the arrays and checking whether each student's interval includes the query time. This brute force method runs in O(n) time and O(1) space, which is optimal for a single query. The logic is simple: count every index where startTime[i] ≤ queryTime ≤ endTime[i].

Is Number of Students Doing Homework at a Given Time asked at Google/Amazon/Meta?

This problem is categorized as an easy array and interval counting problem. Variants of interval overlap counting appear in interviews at companies like Amazon and Google, often with added complexity such as multiple queries or event timelines.

What data structure is used in Number of Students Doing Homework at a Given Time?

The primary data structure is a simple array storing start and end times. The optimized variation may also use sorted arrays with binary search to count events efficiently. No advanced structures like heaps or trees are required for the basic version.

What is the time complexity of Number of Students Doing Homework at a Given Time?

The standard solution runs in O(n) time because you scan each student's start and end time once. Space complexity is O(1) since only a counter variable is required. A binary-search-based variant requires O(n log n) preprocessing due to sorting and O(log n) per query afterward.

Description Hints 2 Solution Video 2:40

1450. Number of Students Doing Homework at a Given Time

Q: Is Number of Students Doing Homework at a Given Time easy or hard?

The problem is rated Easy on LeetCode with a high acceptance rate around 75%. It focuses on understanding interval conditions and simple array iteration rather than complex algorithms.

Q: How to solve Number of Students Doing Homework at a Given Time in O(n)?

Iterate through the startTime and endTime arrays simultaneously. For each student, check if the query time lies between the start and end values inclusive. If it does, increment a counter. After scanning all students, the counter represents how many are doing homework at that moment.

Q: What is the best approach for Number of Students Doing Homework at a Given Time?

The most practical approach is a single pass through the arrays and checking whether each student's interval includes the query time. This brute force method runs in O(n) time and O(1) space, which is optimal for a single query. The logic is simple: count every index where startTime[i] ≤ queryTime ≤ endTime[i].

Q: Is Number of Students Doing Homework at a Given Time asked at Google/Amazon/Meta?

This problem is categorized as an easy array and interval counting problem. Variants of interval overlap counting appear in interviews at companies like Amazon and Google, often with added complexity such as multiple queries or event timelines.

Q: What data structure is used in Number of Students Doing Homework at a Given Time?

The primary data structure is a simple array storing start and end times. The optimized variation may also use sorted arrays with binary search to count events efficiently. No advanced structures like heaps or trees are required for the basic version.

Q: What is the time complexity of Number of Students Doing Homework at a Given Time?

The standard solution runs in O(n) time because you scan each student's start and end time once. Space complexity is O(1) since only a counter variable is required. A binary-search-based variant requires O(n log n) preprocessing due to sorting and O(log n) per query afterward.

Easy76.3% acceptance

Given two integer arrays startTime and endTime and given an integer queryTime.

The ith student started doing their homework at the time startTime[i] and finished it at time endTime[i].

Return the number of students doing their homework at time queryTime. More formally, return the number of students where queryTime lays in the interval [startTime[i], endTime[i]] inclusive.

Example 1:

Input: startTime = [1,2,3], endTime = [3,2,7], queryTime = 4
Output: 1
Explanation: We have 3 students where:
The first student started doing homework at time 1 and finished at time 3 and wasn't doing anything at time 4.
The second student started doing homework at time 2 and finished at time 2 and also wasn't doing anything at time 4.
The third student started doing homework at time 3 and finished at time 7 and was the only student doing homework at time 4.

Example 2:

Input: startTime = [4], endTime = [4], queryTime = 4
Output: 1
Explanation: The only student was doing their homework at the queryTime.

Constraints:

startTime.length == endTime.length
1 <= startTime.length <= 100
1 <= startTime[i] <= endTime[i] <= 1000
1 <= queryTime <= 1000

Topics

Array

Hints open one at a time

Hint 1

Imagine that startTime[i] and endTime[i] form an interval (i.e. [startTime[i], endTime[i]]).

Hint 2

The answer is how many times the queryTime laid in those mentioned intervals.

2:40

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1class Solution:
2    def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,2,3]
[3,2,7]
4

Input

[4]
[4]
4

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