LeetCode - number of students doing homework at a given time

katanazero86·2020년 6월 2일
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number of students doing homework at a given time

  • Difficulty : easy

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.

  • 두 개의 정수 배열 startTime 및 endTime과 정수 queryTime이 제공
  • i 번째 학생은 startTime[i] 에 숙제를 시작하여 endTime[i] 에 숙제를 마침
  • queryTime 시간에 숙제를하는 학생 수를 반환(공식적으로, queryTime이 [startTime [i], endTime [i]] 간격으로 놓인 학생 수를 반환)

exmaple
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

  • startTime.length == endTime.length

solution

  • 작성 언어 : javascript
  • Runtime : 60ms
/**
 * @param {number[]} startTime
 * @param {number[]} endTime
 * @param {number} queryTime
 * @return {number}
 */
var busyStudent = function(startTime, endTime, queryTime) {
    
    const timeLength = startTime.length;
    let busyStudentCount = 0;
    
    for(let i = 0; i < timeLength; i++) {
        if(startTime[i] <= queryTime && endTime[i] >= queryTime) {
            busyStudentCount++;
        }
    }
    
    return busyStudentCount;
    
};
  • 시간복잡도 : O(n)
  • queryTime 이 숙제를 완료해야할 마감시간으로 이해하면 된다.
  • 시작시간 ~ 종료시간 사이에 queryTime 이 포함이 된다면, 숙제 마감을 넘긴 학생
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developer life started : 2016.06.07 ~ 흔한 Front-end 개발자(현재는 백수 얏호👻)

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