README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1450.Number%20of%20Students%20Doing%20Homework%20at%20a%20Given%20Time/README_EN.md	1129	Weekly Contest 189 Q1	Array

1450. Number of Students Doing Homework at a Given Time

中文文档

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 $O(1)$.

Python3

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))

Java

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;
    }
}

C++

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;
    }
};

Go

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

TypeScript

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;
}

Rust

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
    }
}

C

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;
}