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1450. Number of Students Doing Homework at a Given Time

Description

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

Solutions

Solution 1: Direct Traversal

We can directly traverse the two arrays. For each student, we check if \(\textit{queryTime}\) is within their homework time interval. If it is, we increment the answer by one.

After the traversal, we return the answer.

The time complexity is \(O(n)\), where \(n\) is the number of students. The space complexity is \(O(1)\).

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class Solution:
    def busyStudent(
        self, startTime: List[int], endTime: List[int], queryTime: int
    ) -> int:
        return sum(x <= queryTime <= y for x, y in zip(startTime, endTime))

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class Solution {
    public int busyStudent(int[] startTime, int[] endTime, int queryTime) {
        int ans = 0;
        for (int i = 0; i < startTime.length; ++i) {
            if (startTime[i] <= queryTime && queryTime <= endTime[i]) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int busyStudent(vector<int>& startTime, vector<int>& endTime, int queryTime) {
        int ans = 0;
        for (int i = 0; i < startTime.size(); ++i) {
            ans += startTime[i] <= queryTime && queryTime <= endTime[i];
        }
        return ans;
    }
};

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func busyStudent(startTime []int, endTime []int, queryTime int) (ans int) {
    for i, x := range startTime {
        if x <= queryTime && queryTime <= endTime[i] {
            ans++
        }
    }
    return
}

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function busyStudent(startTime: number[], endTime: number[], queryTime: number): number {
    const n = startTime.length;
    let ans = 0;
    for (let i = 0; i < n; i++) {
        if (startTime[i] <= queryTime && queryTime <= endTime[i]) {
            ans++;
        }
    }
    return ans;
}

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impl Solution {
    pub fn busy_student(start_time: Vec<i32>, end_time: Vec<i32>, query_time: i32) -> i32 {
        let mut ans = 0;
        for i in 0..start_time.len() {
            if start_time[i] <= query_time && end_time[i] >= query_time {
                ans += 1;
            }
        }
        ans
    }
}

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

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