leetcode - 2073. Time Needed to Buy Tickets

Description

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

Solution

Brute Force

Just simulation.

Time complexity: $o(sum(A))$
Space complexity: $o(n)$

Math

All numbers before k appear min(A[i], A[k]) times, and all numbers after k appear min(A[i], A[k] - 1) times.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

Brute Force

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        tickets = [(item, index) for index, item in enumerate(tickets)]
        res = 0
        while tickets:
            new_tickets = []
            for item in tickets:
                res += 1
                if item[0] - 1 > 0:
                    new_tickets.append((item[0] - 1, item[1]))
                else:
                    if item[1] == k:
                        return res
            tickets = new_tickets
            if len(tickets) == 1:
                res += tickets[0][0]
                return res

Math

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