time_needed_to_buy.py

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

from typing import List

class Solution:
    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
        ''' Time Needed to Buy Tickets '''
        return sum(min(v, tickets[k] if i <= k else tickets[k] - 1) for i, v in enumerate(tickets))

print(Solution().timeRequiredToBuy([2,3,2], 2))
# Output: 6
print(Solution().timeRequiredToBuy([5,1,1,1], 0))
# Output: 8
print(Solution().timeRequiredToBuy([5, 2, 3, 4], 2))
# Output: 10