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; } }

Todo

print("Todo!")