## Description

https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/

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 = 4Output:1Explanation: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 = 4Output:1Explanation:The only student was doing their homework at the queryTime.

**Example 3:**

Input:startTime = [4], endTime = [4], queryTime = 5Output:0

**Example 4:**

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

**Constraints:**

`startTime.length == endTime.length`

`1 <= startTime.length <= 100`

`1 <= startTime[i] <= endTime[i] <= 1000`

`1 <= queryTime <= 1000`

## Explanation

One pass to check how many qualified intervals are there.

## Python Solution

```
class Solution:
def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:
count = 0
for start, end in zip(startTime, endTime):
if start <= queryTime and queryTime <= end:
count += 1
return count
```

- Time Complexity: O(N).
- Space Complexity: O(1).