Python - Algorithm #41

Python Algorithm

LeetCode

1450. 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.
• Constraints
  • startTime.length == endTime.length
  • 1 ≤ startTime.length ≤ 100
  • 1 ≤ startTime[i] ≤ endTime[i] ≤ 1000
  • 1 ≤ queryTime ≤ 1000

class Solution:
    def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int:
        # for, zip
        answer = 0
        for start, end in zip(startTime, endTime):
            # queryTime이 저 중간에 위치하면 1만큼 answer 늘려주기
            if start <= queryTime <= end:
                answer += 1
        return answer

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

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2073. Time Needed to Buy Tickets

• There are n people in a line queuing to buy tickets, where the 0ᵗʰ person is at the front of the line and the (n - 1)ᵗʰ person is at the back of the line.
• You are given a 0-indexed integer array tickets of length n where the number of tickets that the iᵗʰ person would like to buy is tickets[i].
• Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leave the line.
• Return the time taken for the person at position k (0-indexed) to finish buying tickets.
• Constraints
  • n == tickets.length
  • 1 ≤ n ≤ 100
  • 1 ≤ tickets[i] ≤ 100
  • 0 ≤ k < n

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        # k번째 사람이 사고 싶어 하는 티켓 수와 
        # 특정 위치의 사람이 티켓 사고 싶어 하는 수 비교
        # 해당 위치가 k보다 앞인지 뒤인지
        # 앞에 있는데 사고 싶은 티켓 수가 작은 경우
        # 앞에 있는데 사고 싶은 티켓이 많은 경우
        # 뒤에 있는데 사고 싶은 티켓이 많은 경우
        # 뒤에 있는데 사고 싶은 티켓이 작은 경우
        answer = 0
        for i in range(len(tickets)):
            if i <= k:
                if tickets[i] < tickets[k]:
                    answer += tickets[i]
                else:
                    # 내가 사고 싶은 양을 살 때까지 내 앞에 있음
                    answer += tickets[k]
            else:
                if tickets[i] < tickets[k]:
                    # 자기 사고 싶은 만큼만 사고 떠남
                    answer += tickets[i]
                else:
                    # 내가 사고 싶은 양 살 때까지 줄에 머무는데, 나보다 뒤에 있어서
                    # 내가 다 사게 되는 시점에는 이 사람을 안 기다려도 됨
                    answer += (tickets[k] - 1)
        return answer

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        time = 0
        for i in range(len(tickets)):
            if tickets[i] >= tickets[k]:
                if i <= k:
                    time += tickets[k]
                else:
                    time += tickets[k] - 1
            else:
                time += tickets[i]
        return time

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        n = len(tickets)
        i = 0
        time = 0
        while tickets[k] != 0:
            i %= n
            if tickets[i] != 0:
                tickets[i] -= 1
                time += 1
            i += 1
        return time

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https://leetcode.com/problems/number-of-students-doing-homework-at-a-given-time/

https://leetcode.com/problems/time-needed-to-buy-tickets/