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            "questionId": "2213",
            "questionFrontendId": "2092",
            "categoryTitle": "Algorithms",
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            "title": "Find All People With Secret",
            "titleSlug": "find-all-people-with-secret",
            "content": "<p>You are given an integer <code>n</code> indicating there are <code>n</code> people numbered from <code>0</code> to <code>n - 1</code>. You are also given a <strong>0-indexed</strong> 2D integer array <code>meetings</code> where <code>meetings[i] = [x<sub>i</sub>, y<sub>i</sub>, time<sub>i</sub>]</code> indicates that person <code>x<sub>i</sub></code> and person <code>y<sub>i</sub></code> have a meeting at <code>time<sub>i</sub></code>. A person may attend <strong>multiple meetings</strong> at the same time. Finally, you are given an integer <code>firstPerson</code>.</p>\n\n<p>Person <code>0</code> has a <strong>secret</strong> and initially shares the secret with a person <code>firstPerson</code> at time <code>0</code>. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person <code>x<sub>i</sub></code> has the secret at <code>time<sub>i</sub></code>, then they will share the secret with person <code>y<sub>i</sub></code>, and vice versa.</p>\n\n<p>The secrets are shared <strong>instantaneously</strong>. That is, a person may receive the secret and share it with people in other meetings within the same time frame.</p>\n\n<p>Return <em>a list of all the people that have the secret after all the meetings have taken place. </em>You may return the answer in <strong>any order</strong>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1\n<strong>Output:</strong> [0,1,2,3,5]\n<strong>Explanation:\n</strong>At time 0, person 0 shares the secret with person 1.\nAt time 5, person 1 shares the secret with person 2.\nAt time 8, person 2 shares the secret with person 3.\nAt time 10, person 1 shares the secret with person 5.​​​​\nThus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3\n<strong>Output:</strong> [0,1,3]\n<strong>Explanation:</strong>\nAt time 0, person 0 shares the secret with person 3.\nAt time 2, neither person 1 nor person 2 know the secret.\nAt time 3, person 3 shares the secret with person 0 and person 1.\nThus, people 0, 1, and 3 know the secret after all the meetings.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1\n<strong>Output:</strong> [0,1,2,3,4]\n<strong>Explanation:</strong>\nAt time 0, person 0 shares the secret with person 1.\nAt time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3.\nNote that person 2 can share the secret at the same time as receiving it.\nAt time 2, person 3 shares the secret with person 4.\nThus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= meetings.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>meetings[i].length == 3</code></li>\n\t<li><code>0 &lt;= x<sub>i</sub>, y<sub>i </sub>&lt;= n - 1</code></li>\n\t<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>\n\t<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= firstPerson &lt;= n - 1</code></li>\n</ul>\n",
            "translatedTitle": "找出知晓秘密的所有专家",
            "translatedContent": "<p>给你一个整数 <code>n</code> ，表示有 <code>n</code> 个专家从 <code>0</code> 到 <code>n - 1</code> 编号。另外给你一个下标从 0 开始的二维整数数组 <code>meetings</code> ，其中 <code>meetings[i] = [x<sub>i</sub>, y<sub>i</sub>, time<sub>i</sub>]</code> 表示专家 <code>x<sub>i</sub></code> 和专家 <code>y<sub>i</sub></code> 在时间 <code>time<sub>i</sub></code> 要开一场会。一个专家可以同时参加 <strong>多场会议</strong> 。最后，给你一个整数 <code>firstPerson</code> 。</p>\n\n<p>专家 <code>0</code> 有一个 <strong>秘密</strong> ，最初，他在时间&nbsp;<code>0</code> 将这个秘密分享给了专家 <code>firstPerson</code> 。接着，这个秘密会在每次有知晓这个秘密的专家参加会议时进行传播。更正式的表达是，每次会议，如果专家 <code>x<sub>i</sub></code> 在时间 <code>time<sub>i</sub></code> 时知晓这个秘密，那么他将会与专家 <code>y<sub>i</sub></code> 分享这个秘密，反之亦然。</p>\n\n<p>秘密共享是 <strong>瞬时发生</strong> 的。也就是说，在同一时间，一个专家不光可以接收到秘密，还能在其他会议上与其他专家分享。</p>\n\n<p>在所有会议都结束之后，返回所有知晓这个秘密的专家列表。你可以按 <strong>任何顺序</strong> 返回答案。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1\n<strong>输出：</strong>[0,1,2,3,5]\n<strong>解释：\n</strong>时间 0 ，专家 0 将秘密与专家 1 共享。\n时间 5 ，专家 1 将秘密与专家 2 共享。\n时间 8 ，专家 2 将秘密与专家 3 共享。\n时间 10 ，专家 1 将秘密与专家 5 共享。\n因此，在所有会议结束后，专家 0、1、2、3 和 5 都将知晓这个秘密。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3\n<strong>输出：</strong>[0,1,3]\n<strong>解释：</strong>\n时间 0 ，专家 0 将秘密与专家 3 共享。\n时间 2 ，专家 1 与专家 2 都不知晓这个秘密。\n时间 3 ，专家 3 将秘密与专家 0 和专家 1 共享。\n因此，在所有会议结束后，专家 0、1 和 3 都将知晓这个秘密。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1\n<strong>输出：</strong>[0,1,2,3,4]\n<strong>解释：</strong>\n时间 0 ，专家 0 将秘密与专家 1 共享。\n时间 1 ，专家 1 将秘密与专家 2 共享，专家 2 将秘密与专家 3 共享。\n注意，专家 2 可以在收到秘密的同一时间分享此秘密。\n时间 2 ，专家 3 将秘密与专家 4 共享。\n因此，在所有会议结束后，专家 0、1、2、3 和 4 都将知晓这个秘密。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= meetings.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>meetings[i].length == 3</code></li>\n\t<li><code>0 &lt;= x<sub>i</sub>, y<sub>i </sub>&lt;= n - 1</code></li>\n\t<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>\n\t<li><code>1 &lt;= time<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= firstPerson &lt;= n - 1</code></li>\n</ul>\n",
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            "hints": [
                "Could you model all the meetings happening at the same time as a graph?",
                "What data structure can you use to efficiently share the secret?",
                "You can use the union-find data structure to quickly determine who knows the secret and share the secret."
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href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\\" target=\\\"_blank\\\"> datastructures-js\\/queue<\\/a>\\u3002<\\/p>\"],\"ruby\":[\"Ruby\",\"<p>\\u4f7f\\u7528<code>Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n<p>\\u4e00\\u4e9b\\u5e38\\u7528\\u7684\\u6570\\u636e\\u7ed3\\u6784\\u5df2\\u5728 Algorithms \\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\"<p>\\u7248\\u672c\\uff1a<code>Swift 5.5.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 <a href=\\\"https:\\/\\/swift.org\\/download\\/\\\" target=\\\"_blank\\\">Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\"<p>\\u7248\\u672c\\uff1a<code>Go 1.17<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 <a href=\\\"https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 4.5.4<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.3<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 24.2\"],\"elixir\":[\"Elixir\",\"Elixir 1.13.0 with Erlang\\/OTP 24.2\"]}",
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