Description
https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/
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
Explanation
One pass to check how many qualified intervals are there.
Python Solution
class Solution:
def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:
count = 0
for start, end in zip(startTime, endTime):
if start <= queryTime and queryTime <= end:
count += 1
return count
- Time Complexity: O(N).
- Space Complexity: O(1).