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;
}
}
```