Leetcode - Number of Students Doing Homework at a Given Time Solution
Poor Man
- Published
- Updated
- Reading time
- 2 min read
- Length
- 276 words
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:
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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:
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Input: startTime = [4], endTime = [4], queryTime = 4
Output: 1
Explanation: The only student was doing their homework at the queryTime.Example 3:
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Input: startTime = [4], endTime = [4], queryTime = 5
Output: 0Example 4:
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Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7
Output: 0Example 5:
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Input: startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5
Output: 5Constraints:
startTime.length == endTime.length1 <= startTime.length <= 1001 <= startTime[i] <= endTime[i] <= 10001 <= queryTime <= 1000
Solution in Python#
Python
class Solution:
def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:
return sum(1 for x,y in zip(startTime,endTime) if queryTime>=x and queryTime<=y)