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

# 1450. Number of Students Doing Homework at a Given Time

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 = , endTime = , queryTime = 4

Output: 1

Explanation: The only student was doing their homework at the queryTime.

Example 3:

Input: startTime = , endTime = , 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
}
}