leetcode-problemset/leetcode-cn/originData/first-day-where-you-have-been-in-all-the-rooms.json

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        "question": {
            "questionId": "2124",
            "questionFrontendId": "1997",
            "categoryTitle": "Algorithms",
            "boundTopicId": 972462,
            "title": "First Day Where You Have Been in All the Rooms",
            "titleSlug": "first-day-where-you-have-been-in-all-the-rooms",
            "content": "<p>There are <code>n</code> rooms you need to visit, labeled from <code>0</code> to <code>n - 1</code>. Each day is labeled, starting from <code>0</code>. You will go in and visit one room a day.</p>\n\n<p>Initially on day <code>0</code>, you visit room <code>0</code>. The <strong>order</strong> you visit the rooms for the coming days is determined by the following <strong>rules</strong> and a given <strong>0-indexed</strong> array <code>nextVisit</code> of length <code>n</code>:</p>\n\n<ul>\n\t<li>Assuming that on a day, you visit room <code>i</code>,</li>\n\t<li>if you have been in room <code>i</code> an <strong>odd</strong> number of times (<strong>including</strong> the current visit), on the <strong>next</strong> day you will visit a room with a <strong>lower or equal room number</strong> specified by <code>nextVisit[i]</code> where <code>0 &lt;= nextVisit[i] &lt;= i</code>;</li>\n\t<li>if you have been in room <code>i</code> an <strong>even</strong> number of times (<strong>including</strong> the current visit), on the <strong>next</strong> day you will visit room <code>(i + 1) mod n</code>.</li>\n</ul>\n\n<p>Return <em>the label of the <strong>first</strong> day where you have been in <strong>all</strong> the rooms</em>. It can be shown that such a day exists. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nextVisit = [0,0]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\n- On day 0, you visit room 0. The total times you have been in room 0 is 1, which is odd.\n&nbsp; On the next day you will visit room nextVisit[0] = 0\n- On day 1, you visit room 0, The total times you have been in room 0 is 2, which is even.\n&nbsp; On the next day you will visit room (0 + 1) mod 2 = 1\n- On day 2, you visit room 1. This is the first day where you have been in all the rooms.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nextVisit = [0,0,2]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\nYour room visiting order for each day is: [0,0,1,0,0,1,2,...].\nDay 6 is the first day where you have been in all the rooms.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nextVisit = [0,1,2,0]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\nYour room visiting order for each day is: [0,0,1,1,2,2,3,...].\nDay 6 is the first day where you have been in all the rooms.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nextVisit.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nextVisit[i] &lt;= i</code></li>\n</ul>\n",
            "translatedTitle": "访问完所有房间的第一天",
            "translatedContent": "<p>你需要访问&nbsp;<code>n</code> 个房间，房间从 <code>0</code> 到 <code>n - 1</code> 编号。同时，每一天都有一个日期编号，从 <code>0</code> 开始，依天数递增。你每天都会访问一个房间。</p>\n\n<p>最开始的第 <code>0</code> 天，你访问&nbsp;<code>0</code> 号房间。给你一个长度为 <code>n</code> 且 <strong>下标从 0 开始</strong> 的数组 <code>nextVisit</code> 。在接下来的几天中，你访问房间的 <strong>次序</strong> 将根据下面的 <strong>规则</strong> 决定：</p>\n\n<ul>\n\t<li>假设某一天，你访问&nbsp;<code>i</code> 号房间。</li>\n\t<li>如果算上本次访问，访问&nbsp;<code>i</code> 号房间的次数为 <strong>奇数</strong> ，那么 <strong>第二天</strong> 需要访问&nbsp;<code>nextVisit[i]</code> 所指定的房间，其中 <code>0 &lt;= nextVisit[i] &lt;= i</code> 。</li>\n\t<li>如果算上本次访问，访问&nbsp;<code>i</code> 号房间的次数为 <strong>偶数</strong> ，那么 <strong>第二天</strong> 需要访问&nbsp;<code>(i + 1) mod n</code> 号房间。</li>\n</ul>\n\n<p>请返回你访问完所有房间的第一天的日期编号。题目数据保证总是存在这样的一天。由于答案可能很大，返回对 <code>10<sup>9</sup> + 7</code> 取余后的结果。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nextVisit = [0,0]\n<strong>输出：</strong>2\n<strong>解释：</strong>\n- 第 0 天，你访问房间 0 。访问 0 号房间的总次数为 1 ，次数为奇数。\n&nbsp; 下一天你需要访问房间的编号是 nextVisit[0] = 0\n- 第 1 天，你访问房间 0 。访问 0 号房间的总次数为 2 ，次数为偶数。\n&nbsp; 下一天你需要访问房间的编号是 (0 + 1) mod 2 = 1\n- 第 2 天，你访问房间 1 。这是你第一次完成访问所有房间的那天。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nextVisit = [0,0,2]\n<strong>输出：</strong>6\n<strong>解释：</strong>\n你每天访问房间的次序是 [0,0,1,0,0,1,2,...] 。\n第 6 天是你访问完所有房间的第一天。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nextVisit = [0,1,2,0]\n<strong>输出：</strong>6\n<strong>解释：</strong>\n你每天访问房间的次序是 [0,0,1,1,2,2,3,...] 。\n第 6 天是你访问完所有房间的第一天。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nextVisit.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nextVisit[i] &lt;= i</code></li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int firstDayBeenInAllRooms(vector<int>& nextVisit) {\n\n    }\n};",
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                "The only way to get to room i+1 is when you are visiting room i and room i has been visited an even number of times.",
                "After visiting room i an odd number of times, you are required to visit room nextVisit[i] where nextVisit[i] <= i. It takes a fixed amount of days for you to come back from room nextVisit[i] to room i. Then, you have visited room i even number of times.nextVisit[i]",
                "Can you use Dynamic Programming to avoid recomputing the number of days it takes to visit room i from room nextVisit[i]?"
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