2020-05-16

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

### Question:

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:

Easy two pointers question.

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