[leetcode] 1450. Number of Students Doing Homework at a Given Time

[leetcode] 1450. Number of Students Doing Homework at a Given Time

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 = 4Output: 1Explanation: 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 = 4Output: 1Explanation: The only student was doing their homework at the queryTime.

Example 3:

Input: startTime = [4], endTime = [4], queryTime = 5Output: 0

Example 4:

Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7Output: 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 = 5Output: 5

Constraints:

startTime.length == endTime.length1 <= startTime.length <= 1001 <= startTime[i] <= endTime[i] <= 10001 <= queryTime <= 1000

分析

题目的意思是：给定两个数组，一个表示做作业开始的时间，另一个表示做作业结束的时间，然后给定一个queryTime，判断queryTime时刻正在做作业的人的数量，这里思路也非常直接了。直接遍历数组判断一下就行了，哈哈哈，leetcode居然有这么简单的题目。

代码

class Solution: def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int: res=0 for start,end in zip(startTime,endTime): if(queryTime>=start and queryTime<=end): res+=1 return res

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