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Formatted question description: https://leetcode.ca/all/1450.html

1450. Number of Students Doing Homework at a Given Time

Level

Easy

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.

Example 3:

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

Output: 0

Example 4:

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

Output: 0

Example 5:

Input: startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5

Output: 5

Constraints:

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

Solution

For i from 0 to startTime.length - 1, check whether startTime[i] <= queryTime and endTime[i] >= queryTime. If so, add the counter by 1. Finally, return the counter.

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

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



  • 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;
    }
};

  • func busyStudent(startTime []int, endTime []int, queryTime int) int {
	ans := 0
	for i, a := range startTime {
		b := endTime[i]
		if a <= queryTime && queryTime <= b {
			ans++
		}
	}
	return ans
}

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


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

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