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.length1 <= startTime.length <= 1001 <= startTime[i] <= endTime[i] <= 10001 <= 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;
}
}