LeetCode 2073 Time Needed to Buy Tickets

LeetCode 2073 Time Needed to Buy Tickets

<div>

There are n people in a line queuing to buy tickets, where the 0<sup>th</sup> person is at the front of the line and the (n - 1)<sup>th</sup> 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 i<sup>th</sup> 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:

<pre>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. </pre>

Example 2:

<pre>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. </pre>

Constraints:

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

</div>

题目大意：

排队买票，每个人都有不同的票数需求。每人每次只能买一张，买完后重新排队。买一张票需要1秒，求第k个人买票的总时间。

解题思路：

一开始按照题目要求老老实实每个元素减一，按照流程计算，但效率较低。考虑若所有人票数大于0，每轮计算结果是一样的：
当前排队人数乘以排队的人中的最小票数。当最小票数人离队后，公式会改变。如此循环直到第k个人票数也变成0为止。

解题步骤：

1. 求最小值
2. 计算票数
3. 更新人数，继续循环
4. 结果减去排在第k个人后的人数

注意事项：

1. 结果要减去排在第k个人后的还在排队的人数（tickets数不为负数，可以等于0，因为是同时在同一轮买到足够票）。

Python代码：

def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
	sum, ppl, min_tickets = 0, len(tickets), 0
	while tickets[k] > 0:
		min_tickets = min(t for t in tickets if t > 0)
		sum += min_tickets * ppl
		tickets = [t - min_tickets for t in tickets]
		ppl -= tickets.count(0)

	after_k = [i for i in range(k + 1, len(tickets)) if tickets[i] >= 0]
	return sum - len(after_k)

算法分析：

时间复杂度为O(n)，空间复杂度O(1)。