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Leetcode 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.

Example 1:

Input: tickets = [2,3,2], k = 2
Output: 6
Explanation: 
- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].
- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].
The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.

Example 2:

Input: tickets = [5,1,1,1], k = 0
Output: 8
Explanation:
- In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].
- In the next 4 passes, only the person in position 0 is buying tickets.
The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.

Constraints:

  • n == tickets.length
  • 1 <= n <= 100
  • 1 <= tickets[i] <= 100
  • 0 <= k < n

We can solve this by simple math. Add up the number of tickets needed for each person, the only thing needs to be aware is index if it’s before k then, adding it up if tickets[i] <= tickets[k] else tickets[k]. If it’s after k, then remember to minus 1.

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

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