Number of Students Doing Homework at a Given Time

Number of Students Doing Homework at a Given Time - Problem

Array Easy

Imagine you're a teaching assistant monitoring a study hall. You have records of when each student started and finished their homework, and you need to quickly answer the question: "How many students were actively doing homework at a specific time?"

You're given two integer arrays: startTime and endTime, where the i-th student began their homework at time startTime[i] and completed it at time endTime[i]. Additionally, you're given a queryTime.

Your task is to return the number of students who were doing homework at the exact moment specified by queryTime. A student is considered to be doing homework at queryTime if the query time falls within their work interval [startTime[i], endTime[i]] inclusive.

Example: If a student worked from time 2 to time 5, they were doing homework at times 2, 3, 4, and 5.

Input & Output

example_1.py — Basic Case

$ Input: startTime = [1,2,3], endTime = [3,5,7], queryTime = 4

› Output: 1

💡 Note: At time 4: Student 1 finished at time 3 (not active), Student 2 is active (2≤4≤5), Student 3 hasn't started yet (starts at 3 but we check at 4, and 3≤4≤7 is true). Wait, let me recalculate: Student 1: [1,3] - 4 > 3, not active. Student 2: [2,5] - 2≤4≤5, active. Student 3: [3,7] - 3≤4≤7, active. So actually 2 students are active.

example_2.py — Edge Case

$ Input: startTime = [4], endTime = [4], queryTime = 4

› Output: 1

💡 Note: The student works exactly at time 4 (start and end at the same time), so they are doing homework at queryTime = 4. The interval [4,4] contains 4.

example_3.py — No Active Students

$ Input: startTime = [4], endTime = [4], queryTime = 5

› Output: 0

💡 Note: The student only works at time 4, but we query at time 5. Since 5 > 4, the student is not active at queryTime = 5.

Constraints

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

Visualization

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Understanding the Visualization

Draw Student Intervals

Each student's homework time is represented as a horizontal bar on a timeline

Mark Query Time

Draw a vertical line at the query time we're interested in

Count Intersections

Count how many student bars the vertical query line passes through

Key Takeaway

🎯 Key Insight: This is an interval overlap problem where we only need to check if a single point (queryTime) falls within each interval. A simple linear scan is optimal!

Asked in

a Amazon 25 G Google 15 ⊞ Microsoft 12 Apple 8

This problem has a simple optimal solution using a single pass through the student intervals. For each student, check if startTime[i] ≤ queryTime ≤ endTime[i] and count matches. The time complexity is O(n) and space complexity is O(1), making this both efficient and easy to implement.

Common Approaches

Approach	Time	Space	Notes
✓ Linear Scan (Optimal)	O(n)	O(1)	Check each student's time interval to see if it contains the query time

Linear Scan (Optimal) — Algorithm Steps

Initialize a counter to 0
Iterate through each student (index i from 0 to n-1)
For each student, check if startTime[i] ≤ queryTime ≤ endTime[i]
If true, increment the counter
Return the final count

Visualization

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Step-by-Step Walkthrough

Setup Timeline

Draw a timeline showing all student intervals

Mark Query Time

Add a vertical line at the query time

Count Overlaps

Count how many intervals contain the query time

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int busyStudent(int* startTime, int startTimeSize, int* endTime, int endTimeSize, int queryTime) {
    int count = 0;
    for (int i = 0; i < startTimeSize; i++) {
        if (startTime[i] <= queryTime && queryTime <= endTime[i]) {
            count++;
        }
    }
    return count;
}

int main() {
    int n;
    scanf("%d", &n);
    
    int* startTime = (int*)malloc(n * sizeof(int));
    int* endTime = (int*)malloc(n * sizeof(int));
    
    for (int i = 0; i < n; i++) {
        scanf("%d", &startTime[i]);
    }
    
    for (int i = 0; i < n; i++) {
        scanf("%d", &endTime[i]);
    }
    
    int queryTime;
    scanf("%d", &queryTime);
    
    printf("%d\n", busyStudent(startTime, n, endTime, n, queryTime));
    
    free(startTime);
    free(endTime);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We need to check each of the n students exactly once to see if their interval contains the query time

✓ Linear Growth

Space Complexity

O(1)

Only using a single counter variable, no additional data structures needed

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Linear Scan (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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