leetcode-problemset/leetcode-cn/originData/time-needed-to-buy-tickets.json

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        "question": {
            "questionId": "2195",
            "questionFrontendId": "2073",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1097162,
            "title": "Time Needed to Buy Tickets",
            "titleSlug": "time-needed-to-buy-tickets",
            "content": "<p>There are <code>n</code> people in a line queuing to buy tickets, where the <code>0<sup>th</sup></code> person is at the <strong>front</strong> of the line and the <code>(n - 1)<sup>th</sup></code> person is at the <strong>back</strong> of the line.</p>\n\n<p>You are given a <strong>0-indexed</strong> integer array <code>tickets</code> of length <code>n</code> where the number of tickets that the <code>i<sup>th</sup></code> person would like to buy is <code>tickets[i]</code>.</p>\n\n<p>Each person takes <strong>exactly 1 second</strong> to buy a ticket. A person can only buy <strong>1 ticket at a time</strong> and has to go back to <strong>the end</strong> of the line (which happens <strong>instantaneously</strong>) in order to buy more tickets. If a person does not have any tickets left to buy, the person will <strong>leave </strong>the line.</p>\n\n<p>Return <em>the <strong>time taken</strong> for the person at position </em><code>k</code><em>&nbsp;</em><strong><em>(0-indexed)</em>&nbsp;</strong><em>to finish buying tickets</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> tickets = [2,3,2], k = 2\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> \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&nbsp;position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> tickets = [5,1,1,1], k = 0\n<strong>Output:</strong> 8\n<strong>Explanation:</strong>\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&nbsp;position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == tickets.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 100</code></li>\n\t<li><code>1 &lt;= tickets[i] &lt;= 100</code></li>\n\t<li><code>0 &lt;= k &lt; n</code></li>\n</ul>\n",
            "translatedTitle": "买票需要的时间",
            "translatedContent": "<p>有 <code>n</code> 个人前来排队买票，其中第 <code>0</code> 人站在队伍 <strong>最前方</strong> ，第 <code>(n - 1)</code> 人站在队伍 <strong>最后方</strong> 。</p>\n\n<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>tickets</code> ，数组长度为 <code>n</code> ，其中第 <code>i</code> 人想要购买的票数为 <code>tickets[i]</code> 。</p>\n\n<p>每个人买票都需要用掉 <strong>恰好 1 秒</strong> 。一个人 <strong>一次只能买一张票</strong> ，如果需要购买更多票，他必须走到&nbsp; <strong>队尾</strong> 重新排队（<strong>瞬间 </strong>发生，不计时间）。如果一个人没有剩下需要买的票，那他将会 <strong>离开</strong> 队伍。</p>\n\n<p>返回位于位置 <code>k</code>（下标从 <strong>0</strong> 开始）的人完成买票需要的时间（以秒为单位）。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>tickets = [2,3,2], k = 2\n<strong>输出：</strong>6\n<strong>解释：</strong> \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</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>tickets = [5,1,1,1], k = 0\n<strong>输出：</strong>8\n<strong>解释：</strong>\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</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == tickets.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 100</code></li>\n\t<li><code>1 &lt;= tickets[i] &lt;= 100</code></li>\n\t<li><code>0 &lt;= k &lt; n</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int timeRequiredToBuy(vector<int>& tickets, int k) {\n        \n    }\n};",
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                    "code": "/**\n * @param {number[]} tickets\n * @param {number} k\n * @return {number}\n */\nvar timeRequiredToBuy = function(tickets, k) {\n\n};",
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                "Loop through the line of people and decrement the number of tickets for each to buy one at a time as if simulating the line moving forward. Keep track of how many tickets have been sold up until person k has no more tickets to buy.",
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