LeetCode 1450. Number of Students Doing Homework at a Given Time

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).

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