mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
206 lines
30 KiB
JSON
206 lines
30 KiB
JSON
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"questionId": "2512",
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"questionFrontendId": "2424",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1859568,
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"title": "Longest Uploaded Prefix",
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"titleSlug": "longest-uploaded-prefix",
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"content": "<p>You are given a stream of <code>n</code> videos, each represented by a <strong>distinct</strong> number from <code>1</code> to <code>n</code> that you need to "upload" to a server. You need to implement a data structure that calculates the length of the <strong>longest uploaded prefix</strong> at various points in the upload process.</p>\n\n<p>We consider <code>i</code> to be an uploaded prefix if all videos in the range <code>1</code> to <code>i</code> (<strong>inclusive</strong>) have been uploaded to the server. The longest uploaded prefix is the <strong>maximum </strong>value of <code>i</code> that satisfies this definition.<br />\n<br />\nImplement the <code>LUPrefix </code>class:</p>\n\n<ul>\n\t<li><code>LUPrefix(int n)</code> Initializes the object for a stream of <code>n</code> videos.</li>\n\t<li><code>void upload(int video)</code> Uploads <code>video</code> to the server.</li>\n\t<li><code>int longest()</code> Returns the length of the <strong>longest uploaded prefix</strong> defined above.</li>\n</ul>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n["LUPrefix", "upload", "longest", "upload", "longest", "upload", "longest"]\n[[4], [3], [], [1], [], [2], []]\n<strong>Output</strong>\n[null, null, 0, null, 1, null, 3]\n\n<strong>Explanation</strong>\nLUPrefix server = new LUPrefix(4); // Initialize a stream of 4 videos.\nserver.upload(3); // Upload video 3.\nserver.longest(); // Since video 1 has not been uploaded yet, there is no prefix.\n // So, we return 0.\nserver.upload(1); // Upload video 1.\nserver.longest(); // The prefix [1] is the longest uploaded prefix, so we return 1.\nserver.upload(2); // Upload video 2.\nserver.longest(); // The prefix [1,2,3] is the longest uploaded prefix, so we return 3.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= video <= n</code></li>\n\t<li>All values of <code>video</code> are <strong>distinct</strong>.</li>\n\t<li>At most <code>2 * 10<sup>5</sup></code> calls <strong>in total</strong> will be made to <code>upload</code> and <code>longest</code>.</li>\n\t<li>At least one call will be made to <code>longest</code>.</li>\n</ul>\n",
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"translatedTitle": "最长上传前缀",
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"translatedContent": "<p>给你一个 <code>n</code> 个视频的上传序列，每个视频编号为 <code>1</code> 到 <code>n</code> 之间的 <strong>不同</strong> 数字，你需要依次将这些视频上传到服务器。请你实现一个数据结构，在上传的过程中计算 <strong>最长上传前缀</strong> 。</p>\n\n<p>如果 <strong>闭区间</strong> <code>1</code> 到 <code>i</code> 之间的视频全部都已经被上传到服务器，那么我们称 <code>i</code> 是上传前缀。最长上传前缀指的是符合定义的 <code>i</code> 中的 <strong>最大值</strong> 。<br>\n<br>\n请你实现 <code>LUPrefix</code> 类：</p>\n\n<ul>\n\t<li><code>LUPrefix(int n)</code> 初始化一个 <code>n</code> 个视频的流对象。</li>\n\t<li><code>void upload(int video)</code> 上传 <code>video</code> 到服务器。</li>\n\t<li><code>int longest()</code> 返回上述定义的 <strong>最长上传前缀</strong> 的长度。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>\n[\"LUPrefix\", \"upload\", \"longest\", \"upload\", \"longest\", \"upload\", \"longest\"]\n[[4], [3], [], [1], [], [2], []]\n<strong>输出：</strong>\n[null, null, 0, null, 1, null, 3]\n\n<strong>解释：</strong>\nLUPrefix server = new LUPrefix(4); // 初始化 4个视频的上传流\nserver.upload(3); // 上传视频 3 。\nserver.longest(); // 由于视频 1 还没有被上传，最长上传前缀是 0 。\nserver.upload(1); // 上传视频 1 。\nserver.longest(); // 前缀 [1] 是最长上传前缀，所以我们返回 1 。\nserver.upload(2); // 上传视频 2 。\nserver.longest(); // 前缀 [1,2,3] 是最长上传前缀，所以我们返回 3 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= video <= 10<sup>5</sup></code></li>\n\t<li><code>video</code> 中所有值 <strong>互不相同</strong> 。</li>\n\t<li><code>upload</code> 和 <code>longest</code> <strong>总调用</strong> 次数至多不超过 <code>2 * 10<sup>5</sup></code> 次。</li>\n\t<li>至少会调用 <code>longest</code> 一次。</li>\n</ul>\n",
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"code": "class LUPrefix {\npublic:\n LUPrefix(int n) {\n\n }\n \n void upload(int video) {\n\n }\n \n int longest() {\n\n }\n};\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * LUPrefix* obj = new LUPrefix(n);\n * obj->upload(video);\n * int param_2 = obj->longest();\n */",
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"code": "class LUPrefix {\n\n public LUPrefix(int n) {\n\n }\n \n public void upload(int video) {\n\n }\n \n public int longest() {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * LUPrefix obj = new LUPrefix(n);\n * obj.upload(video);\n * int param_2 = obj.longest();\n */",
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"code": "class LUPrefix(object):\n\n def __init__(self, n):\n \"\"\"\n :type n: int\n \"\"\"\n\n\n def upload(self, video):\n \"\"\"\n :type video: int\n :rtype: None\n \"\"\"\n\n\n def longest(self):\n \"\"\"\n :rtype: int\n \"\"\"\n\n\n\n# Your LUPrefix object will be instantiated and called as such:\n# obj = LUPrefix(n)\n# obj.upload(video)\n# param_2 = obj.longest()",
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"code": "class LUPrefix:\n\n def __init__(self, n: int):\n\n\n def upload(self, video: int) -> None:\n\n\n def longest(self) -> int:\n\n\n\n# Your LUPrefix object will be instantiated and called as such:\n# obj = LUPrefix(n)\n# obj.upload(video)\n# param_2 = obj.longest()",
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"code": "\n\n\ntypedef struct {\n\n} LUPrefix;\n\n\nLUPrefix* lUPrefixCreate(int n) {\n\n}\n\nvoid lUPrefixUpload(LUPrefix* obj, int video) {\n\n}\n\nint lUPrefixLongest(LUPrefix* obj) {\n\n}\n\nvoid lUPrefixFree(LUPrefix* obj) {\n\n}\n\n/**\n * Your LUPrefix struct will be instantiated and called as such:\n * LUPrefix* obj = lUPrefixCreate(n);\n * lUPrefixUpload(obj, video);\n \n * int param_2 = lUPrefixLongest(obj);\n \n * lUPrefixFree(obj);\n*/",
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"code": "public class LUPrefix {\n\n public LUPrefix(int n) {\n\n }\n \n public void Upload(int video) {\n\n }\n \n public int Longest() {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * LUPrefix obj = new LUPrefix(n);\n * obj.Upload(video);\n * int param_2 = obj.Longest();\n */",
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"code": "/**\n * @param {number} n\n */\nvar LUPrefix = function(n) {\n\n};\n\n/** \n * @param {number} video\n * @return {void}\n */\nLUPrefix.prototype.upload = function(video) {\n\n};\n\n/**\n * @return {number}\n */\nLUPrefix.prototype.longest = function() {\n\n};\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * var obj = new LUPrefix(n)\n * obj.upload(video)\n * var param_2 = obj.longest()\n */",
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"code": "class LUPrefix\n\n=begin\n :type n: Integer\n=end\n def initialize(n)\n\n end\n\n\n=begin\n :type video: Integer\n :rtype: Void\n=end\n def upload(video)\n\n end\n\n\n=begin\n :rtype: Integer\n=end\n def longest()\n\n end\n\n\nend\n\n# Your LUPrefix object will be instantiated and called as such:\n# obj = LUPrefix.new(n)\n# obj.upload(video)\n# param_2 = obj.longest()",
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"code": "\nclass LUPrefix {\n\n init(_ n: Int) {\n\n }\n \n func upload(_ video: Int) {\n\n }\n \n func longest() -> Int {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * let obj = LUPrefix(n)\n * obj.upload(video)\n * let ret_2: Int = obj.longest()\n */",
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"code": "type LUPrefix struct {\n\n}\n\n\nfunc Constructor(n int) LUPrefix {\n\n}\n\n\nfunc (this *LUPrefix) Upload(video int) {\n\n}\n\n\nfunc (this *LUPrefix) Longest() int {\n\n}\n\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * obj := Constructor(n);\n * obj.Upload(video);\n * param_2 := obj.Longest();\n */",
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"code": "class LUPrefix(_n: Int) {\n\n def upload(video: Int) {\n\n }\n\n def longest(): Int = {\n\n }\n\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * var obj = new LUPrefix(n)\n * obj.upload(video)\n * var param_2 = obj.longest()\n */",
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"lang": "Kotlin",
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"code": "class LUPrefix(n: Int) {\n\n fun upload(video: Int) {\n\n }\n\n fun longest(): Int {\n\n }\n\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * var obj = LUPrefix(n)\n * obj.upload(video)\n * var param_2 = obj.longest()\n */",
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"code": "struct LUPrefix {\n\n}\n\n\n/**\n * `&self` means the method takes an immutable reference.\n * If you need a mutable reference, change it to `&mut self` instead.\n */\nimpl LUPrefix {\n\n fn new(n: i32) -> Self {\n\n }\n \n fn upload(&self, video: i32) {\n\n }\n \n fn longest(&self) -> i32 {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * let obj = LUPrefix::new(n);\n * obj.upload(video);\n * let ret_2: i32 = obj.longest();\n */",
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"code": "class LUPrefix {\n /**\n * @param Integer $n\n */\n function __construct($n) {\n\n }\n\n /**\n * @param Integer $video\n * @return NULL\n */\n function upload($video) {\n\n }\n\n /**\n * @return Integer\n */\n function longest() {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * $obj = LUPrefix($n);\n * $obj->upload($video);\n * $ret_2 = $obj->longest();\n */",
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"code": "class LUPrefix {\n constructor(n: number) {\n\n }\n\n upload(video: number): void {\n\n }\n\n longest(): number {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * var obj = new LUPrefix(n)\n * obj.upload(video)\n * var param_2 = obj.longest()\n */",
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"lang": "Racket",
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"langSlug": "racket",
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"code": "(define lu-prefix%\n (class object%\n (super-new)\n \n ; n : exact-integer?\n (init-field\n n)\n \n ; upload : exact-integer? -> void?\n (define/public (upload video)\n\n )\n ; longest : -> exact-integer?\n (define/public (longest)\n\n )))\n\n;; Your lu-prefix% object will be instantiated and called as such:\n;; (define obj (new lu-prefix% [n n]))\n;; (send obj upload video)\n;; (define param_2 (send obj longest))",
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"code": "-spec lu_prefix_init_(N :: integer()) -> any().\nlu_prefix_init_(N) ->\n .\n\n-spec lu_prefix_upload(Video :: integer()) -> any().\nlu_prefix_upload(Video) ->\n .\n\n-spec lu_prefix_longest() -> integer().\nlu_prefix_longest() ->\n .\n\n\n%% Your functions will be called as such:\n%% lu_prefix_init_(N),\n%% lu_prefix_upload(Video),\n%% Param_2 = lu_prefix_longest(),\n\n%% lu_prefix_init_ will be called before every test case, in which you can do some necessary initializations.",
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"code": "defmodule LUPrefix do\n @spec init_(n :: integer) :: any\n def init_(n) do\n\n end\n\n @spec upload(video :: integer) :: any\n def upload(video) do\n\n end\n\n @spec longest() :: integer\n def longest() do\n\n end\nend\n\n# Your functions will be called as such:\n# LUPrefix.init_(n)\n# LUPrefix.upload(video)\n# param_2 = LUPrefix.longest()\n\n# LUPrefix.init_ will be called before every test case, in which you can do some necessary initializations.",
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"code": "class LUPrefix {\n\n LUPrefix(int n) {\n\n }\n \n void upload(int video) {\n\n }\n \n int longest() {\n\n }\n}\n\n/**\n * Your LUPrefix object will be instantiated and called as such:\n * LUPrefix obj = LUPrefix(n);\n * obj.upload(video);\n * int param2 = obj.longest();\n */",
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"Maintain an array keeping track of whether video “i” has been uploaded yet."
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