{ "data": { "question": { "questionId": "2195", "questionFrontendId": "2073", "categoryTitle": "Algorithms", "boundTopicId": 1097162, "title": "Time Needed to Buy Tickets", "titleSlug": "time-needed-to-buy-tickets", "content": "
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.
\n\nReturn the time taken for the person at position k
(0-indexed) to finish buying tickets.
\n
Example 1:
\n\n\nInput: tickets = [2,3,2], k = 2\nOutput: 6\nExplanation: \n- In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1].\n- In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0].\nThe person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.\n\n\n
Example 2:
\n\n\nInput: tickets = [5,1,1,1], k = 0\nOutput: 8\nExplanation:\n- In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0].\n- In the next 4 passes, only the person in position 0 is buying tickets.\nThe person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.\n\n\n
\n
Constraints:
\n\nn == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n
有 n
个人前来排队买票,其中第 0
人站在队伍 最前方 ,第 (n - 1)
人站在队伍 最后方 。
给你一个下标从 0 开始的整数数组 tickets
,数组长度为 n
,其中第 i
人想要购买的票数为 tickets[i]
。
每个人买票都需要用掉 恰好 1 秒 。一个人 一次只能买一张票 ,如果需要购买更多票,他必须走到 队尾 重新排队(瞬间 发生,不计时间)。如果一个人没有剩下需要买的票,那他将会 离开 队伍。
\n\n返回位于位置 k
(下标从 0 开始)的人完成买票需要的时间(以秒为单位)。
\n\n
示例 1:
\n\n输入:tickets = [2,3,2], k = 2\n输出:6\n解释: \n- 第一轮,队伍中的每个人都买到一张票,队伍变为 [1, 2, 1] 。\n- 第二轮,队伍中的每个都又都买到一张票,队伍变为 [0, 1, 0] 。\n位置 2 的人成功买到 2 张票,用掉 3 + 3 = 6 秒。\n\n\n
示例 2:
\n\n输入:tickets = [5,1,1,1], k = 0\n输出:8\n解释:\n- 第一轮,队伍中的每个人都买到一张票,队伍变为 [4, 0, 0, 0] 。\n- 接下来的 4 轮,只有位置 0 的人在买票。\n位置 0 的人成功买到 5 张票,用掉 4 + 1 + 1 + 1 + 1 = 8 秒。\n\n\n
\n\n
提示:
\n\nn == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n
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