Given two integer arrays
endTime and given an integer
ith student started doing their homework at the time
startTime[i] and finished it at time
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.
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.
Input: startTime = , endTime = , queryTime = 4 Output: 1 Explanation: The only student was doing their homework at the queryTime.
Input: startTime = , endTime = , queryTime = 5 Output: 0
Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7 Output: 0
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
startTime.length == endTime.length
1 <= startTime.length <= 100
1 <= startTime[i] <= endTime[i] <= 1000
1 <= queryTime <= 1000
One pass to check how many qualified intervals are there.
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).