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;
}
}

• Todo

• print("Todo!")