Time Needed to Buy Tickets Python Solution

lcpy gen -n 2073   # by problem number
lcpy gen -s time_needed_to_buy_tickets   # by problem name

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Problem

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Examples

Input: tickets = [2,3,2], k = 2
Output: 6

• The queue starts as [2,3,2], where the kth person is underlined.
• After the person at the front has bought a ticket, the queue becomes [3,2,1] at 1 second.
• Continuing this process, the queue becomes [2,1,2] at 2 seconds.
• Continuing this process, the queue becomes [1,2,1] at 3 seconds.
• Continuing this process, the queue becomes [2,1] at 4 seconds. Note: the person at the front left the queue.
• Continuing this process, the queue becomes [1,1] at 5 seconds.
• Continuing this process, the queue becomes [1] at 6 seconds. The kth person has bought all their tickets, so return 6.

Input: tickets = [5,1,1,1], k = 0
Output: 8

• The queue starts as [5,1,1,1], where the kth person is underlined.
• After the person at the front has bought a ticket, the queue becomes [1,1,1,4] at 1 second.
• Continuing this process for 3 seconds, the queue becomes [4] at 4 seconds.
• Continuing this process for 4 seconds, the queue becomes [] at 8 seconds. The kth person has bought all their tickets, so return 8.

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Constraints

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

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Solution

class Solution:
    # Time: O(n)
    # Space: O(1)
    def time_required_to_buy(self, tickets: list[int], k: int) -> int:
        target = tickets[k]
        total = 0
        for i, need in enumerate(tickets):
            if i <= k:
                total += min(need, target)
            else:
                total += min(need, target - 1)
        return total

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Complexity

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