leetcode-problemset/leetcode-cn/originData/count-k-subsequences-of-a-string-with-maximum-beauty.json

{
    "data": {
        "question": {
            "questionId": "3057",
            "questionFrontendId": "2842",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2421034,
            "title": "Count K-Subsequences of a String With Maximum Beauty",
            "titleSlug": "count-k-subsequences-of-a-string-with-maximum-beauty",
            "content": "<p>You are given a string <code>s</code> and an integer <code>k</code>.</p>\n\n<p>A <strong>k-subsequence</strong> is a <strong>subsequence</strong> of <code>s</code>, having length <code>k</code>, and all its characters are <strong>unique</strong>, <strong>i.e</strong>., every character occurs once.</p>\n\n<p>Let <code>f(c)</code> denote the number of times the character <code>c</code> occurs in <code>s</code>.</p>\n\n<p>The <strong>beauty</strong> of a <strong>k-subsequence</strong> is the <strong>sum</strong> of <code>f(c)</code> for every character <code>c</code> in the k-subsequence.</p>\n\n<p>For example, consider <code>s = &quot;abbbdd&quot;</code> and <code>k = 2</code>:</p>\n\n<ul>\n\t<li><code>f(&#39;a&#39;) = 1</code>, <code>f(&#39;b&#39;) = 3</code>, <code>f(&#39;d&#39;) = 2</code></li>\n\t<li>Some k-subsequences of <code>s</code> are:\n\t<ul>\n\t\t<li><code>&quot;<u><strong>ab</strong></u>bbdd&quot;</code> -&gt; <code>&quot;ab&quot;</code> having a beauty of <code>f(&#39;a&#39;) + f(&#39;b&#39;) = 4</code></li>\n\t\t<li><code>&quot;<u><strong>a</strong></u>bbb<strong><u>d</u></strong>d&quot;</code> -&gt; <code>&quot;ad&quot;</code> having a beauty of <code>f(&#39;a&#39;) + f(&#39;d&#39;) = 3</code></li>\n\t\t<li><code>&quot;a<strong><u>b</u></strong>bb<u><strong>d</strong></u>d&quot;</code> -&gt; <code>&quot;bd&quot;</code> having a beauty of <code>f(&#39;b&#39;) + f(&#39;d&#39;) = 5</code></li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>Return <em>an integer denoting the number of k-subsequences </em><em>whose <strong>beauty</strong> is the <strong>maximum</strong> among all <strong>k-subsequences</strong></em>. Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>A subsequence of a string is a new string formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.</p>\n\n<p><strong>Notes</strong></p>\n\n<ul>\n\t<li><code>f(c)</code> is the number of times a character <code>c</code> occurs in <code>s</code>, not a k-subsequence.</li>\n\t<li>Two k-subsequences are considered different if one is formed by an index that is not present in the other. So, two k-subsequences may form the same string.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;bcca&quot;, k = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> <span style=\"white-space: normal\">From s we have f(&#39;a&#39;) = 1, f(&#39;b&#39;) = 1, and f(&#39;c&#39;) = 2.</span>\nThe k-subsequences of s are: \n<strong><u>bc</u></strong>ca having a beauty of f(&#39;b&#39;) + f(&#39;c&#39;) = 3 \n<strong><u>b</u></strong>c<u><strong>c</strong></u>a having a beauty of f(&#39;b&#39;) + f(&#39;c&#39;) = 3 \n<strong><u>b</u></strong>cc<strong><u>a</u></strong> having a beauty of f(&#39;b&#39;) + f(&#39;a&#39;) = 2 \nb<strong><u>c</u></strong>c<u><strong>a</strong></u><strong> </strong>having a beauty of f(&#39;c&#39;) + f(&#39;a&#39;) = 3\nbc<strong><u>ca</u></strong> having a beauty of f(&#39;c&#39;) + f(&#39;a&#39;) = 3 \nThere are 4 k-subsequences that have the maximum beauty, 3. \nHence, the answer is 4. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abbcd&quot;, k = 4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> From s we have f(&#39;a&#39;) = 1, f(&#39;b&#39;) = 2, f(&#39;c&#39;) = 1, and f(&#39;d&#39;) = 1. \nThe k-subsequences of s are: \n<u><strong>ab</strong></u>b<strong><u>cd</u></strong> having a beauty of f(&#39;a&#39;) + f(&#39;b&#39;) + f(&#39;c&#39;) + f(&#39;d&#39;) = 5\n<u style=\"white-space: normal;\"><strong>a</strong></u>b<u><strong>bcd</strong></u> having a beauty of f(&#39;a&#39;) + f(&#39;b&#39;) + f(&#39;c&#39;) + f(&#39;d&#39;) = 5 \nThere are 2 k-subsequences that have the maximum beauty, 5. \nHence, the answer is 2. \n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= s.length</code></li>\n\t<li><code>s</code> consists only of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "统计一个字符串的 k 子序列美丽值最大的数目",
            "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;。</p>\n\n<p><strong>k 子序列</strong>指的是 <code>s</code>&nbsp;的一个长度为 <code>k</code>&nbsp;的 <strong>子序列</strong>&nbsp;，且所有字符都是 <strong>唯一</strong>&nbsp;的，也就是说每个字符在子序列里只出现过一次。</p>\n\n<p>定义&nbsp;<code>f(c)</code>&nbsp;为字符 <code>c</code>&nbsp;在 <code>s</code>&nbsp;中出现的次数。</p>\n\n<p>k 子序列的 <strong>美丽值</strong>&nbsp;定义为这个子序列中每一个字符 <code>c</code>&nbsp;的&nbsp;<code>f(c)</code>&nbsp;之 <strong>和</strong>&nbsp;。</p>\n\n<p>比方说，<code>s = \"abbbdd\"</code>&nbsp;和&nbsp;<code>k = 2</code>&nbsp;，我们有：</p>\n\n<ul>\n\t<li><code>f('a') = 1</code>, <code>f('b') = 3</code>, <code>f('d') = 2</code></li>\n\t<li><code>s</code>&nbsp;的部分 k 子序列为：\n\t<ul>\n\t\t<li><code>\"<em><strong>ab</strong></em>bbdd\"</code> -&gt; <code>\"ab\"</code>&nbsp;，美丽值为&nbsp;<code>f('a') + f('b') = 4</code></li>\n\t\t<li><code>\"<em><strong>a</strong></em>bbb<em><strong>d</strong></em>d\"</code> -&gt; <code>\"ad\"</code>&nbsp;，美丽值为&nbsp;<code>f('a') + f('d') = 3</code></li>\n\t\t<li><code>\"a<em><strong>b</strong></em>bb<em><strong>d</strong></em>d\"</code> -&gt; <code>\"bd\"</code>&nbsp;，美丽值为&nbsp;<code>f('b') + f('d') = 5</code></li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>请你返回一个整数，表示所有 <strong>k 子序列&nbsp;</strong>里面 <strong>美丽值 </strong>是&nbsp;<strong>最大值</strong>&nbsp;的子序列数目。由于答案可能很大，将结果对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;取余后返回。</p>\n\n<p>一个字符串的子序列指的是从原字符串里面删除一些字符（也可能一个字符也不删除），不改变剩下字符顺序连接得到的新字符串。</p>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li><code>f(c)</code> 指的是字符&nbsp;<code>c</code>&nbsp;在字符串&nbsp;<code>s</code>&nbsp;的出现次数，不是在 k 子序列里的出现次数。</li>\n\t<li>两个 k 子序列如果有任何一个字符在原字符串中的下标不同，则它们是两个不同的子序列。所以两个不同的 k 子序列可能产生相同的字符串。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"bcca\", k = 2\n<b>输出：</b>4\n<b>解释：</b><span style=\"white-space: normal\">s 中我们有 f('a') = 1 ，f('b') = 1 和 f('c') = 2 。</span>\ns 的 k 子序列为：\n<em><strong>bc</strong></em>ca ，美丽值为 f('b') + f('c') = 3\n<em><strong>b</strong></em>c<em><strong>c</strong></em>a ，美丽值为 f('b') + f('c') = 3\n<em><strong>b</strong></em>cc<em><strong>a</strong></em> ，美丽值为 f('b') + f('a') = 2\nb<em><strong>c</strong></em>c<em><strong>a</strong></em><strong> </strong>，美丽值为 f('c') + f('a') = 3\nbc<em><strong>ca</strong></em> ，美丽值为 f('c') + f('a') = 3\n总共有 4 个 k 子序列美丽值为最大值 3 。\n所以答案为 4 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"abbcd\", k = 4\n<b>输出：</b>2\n<b>解释：</b><span style=\"white-space: normal\">s 中我们有 f('a') = 1 ，f('b') = 2&nbsp;，f('c') = 1&nbsp;和</span> f('d') = 1 。\ns 的 k 子序列为：\n<em><strong>ab</strong></em>b<em><strong>cd</strong></em> ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5\n<span style=\"white-space: normal;\"><b><i>a</i></b></span>b<em><strong>bcd</strong></em> ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5 \n总共有 2 个 k 子序列美丽值为最大值 5 。\n所以答案为 2 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= k &lt;= s.length</code></li>\n\t<li><code>s</code>&nbsp;只包含小写英文字母。</li>\n</ul>\n",
            "isPaidOnly": false,
            "difficulty": "Hard",
            "likes": 7,
            "dislikes": 0,
            "isLiked": null,
            "similarQuestions": "[]",
            "contributors": [],
            "langToValidPlayground": "{\"cpp\": true, \"java\": true, \"python\": true, \"python3\": true, \"mysql\": false, \"mssql\": false, \"oraclesql\": false, \"c\": false, \"csharp\": false, \"javascript\": false, \"typescript\": false, \"bash\": false, \"php\": false, \"swift\": false, \"kotlin\": false, \"dart\": false, \"golang\": false, \"ruby\": false, \"scala\": false, \"html\": false, \"pythonml\": false, \"rust\": false, \"racket\": false, \"erlang\": false, \"elixir\": false, \"pythondata\": false, \"react\": false, \"vanillajs\": false, \"postgresql\": false}",
            "topicTags": [
                {
                    "name": "Greedy",
                    "slug": "greedy",
                    "translatedName": "贪心",
                    "__typename": "TopicTagNode"
                },
                {
                    "name": "Hash Table",
                    "slug": "hash-table",
                    "translatedName": "哈希表",
                    "__typename": "TopicTagNode"
                },
                {
                    "name": "Math",
                    "slug": "math",
                    "translatedName": "数学",
                    "__typename": "TopicTagNode"
                },
                {
                    "name": "String",
                    "slug": "string",
                    "translatedName": "字符串",
                    "__typename": "TopicTagNode"
                },
                {
                    "name": "Combinatorics",
                    "slug": "combinatorics",
                    "translatedName": "组合数学",
                    "__typename": "TopicTagNode"
                }
            ],
            "companyTagStats": null,
            "codeSnippets": [
                {
                    "lang": "C++",
                    "langSlug": "cpp",
                    "code": "class Solution {\npublic:\n    int countKSubsequencesWithMaxBeauty(string s, int k) {\n        \n    }\n};",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Java",
                    "langSlug": "java",
                    "code": "class Solution {\n    public int countKSubsequencesWithMaxBeauty(String s, int k) {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Python",
                    "langSlug": "python",
                    "code": "class Solution(object):\n    def countKSubsequencesWithMaxBeauty(self, s, k):\n        \"\"\"\n        :type s: str\n        :type k: int\n        :rtype: int\n        \"\"\"",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Python3",
                    "langSlug": "python3",
                    "code": "class Solution:\n    def countKSubsequencesWithMaxBeauty(self, s: str, k: int) -> int:",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "C",
                    "langSlug": "c",
                    "code": "int countKSubsequencesWithMaxBeauty(char* s, int k){\n\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "C#",
                    "langSlug": "csharp",
                    "code": "public class Solution {\n    public int CountKSubsequencesWithMaxBeauty(string s, int k) {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "JavaScript",
                    "langSlug": "javascript",
                    "code": "/**\n * @param {string} s\n * @param {number} k\n * @return {number}\n */\nvar countKSubsequencesWithMaxBeauty = function(s, k) {\n\n};",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "TypeScript",
                    "langSlug": "typescript",
                    "code": "function countKSubsequencesWithMaxBeauty(s: string, k: number): number {\n\n};",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "PHP",
                    "langSlug": "php",
                    "code": "class Solution {\n\n    /**\n     * @param String $s\n     * @param Integer $k\n     * @return Integer\n     */\n    function countKSubsequencesWithMaxBeauty($s, $k) {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Swift",
                    "langSlug": "swift",
                    "code": "class Solution {\n    func countKSubsequencesWithMaxBeauty(_ s: String, _ k: Int) -> Int {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Kotlin",
                    "langSlug": "kotlin",
                    "code": "class Solution {\n    fun countKSubsequencesWithMaxBeauty(s: String, k: Int): Int {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Dart",
                    "langSlug": "dart",
                    "code": "class Solution {\n  int countKSubsequencesWithMaxBeauty(String s, int k) {\n\n  }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Go",
                    "langSlug": "golang",
                    "code": "func countKSubsequencesWithMaxBeauty(s string, k int) int {\n\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Ruby",
                    "langSlug": "ruby",
                    "code": "# @param {String} s\n# @param {Integer} k\n# @return {Integer}\ndef count_k_subsequences_with_max_beauty(s, k)\n\nend",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Scala",
                    "langSlug": "scala",
                    "code": "object Solution {\n    def countKSubsequencesWithMaxBeauty(s: String, k: Int): Int = {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Rust",
                    "langSlug": "rust",
                    "code": "impl Solution {\n    pub fn count_k_subsequences_with_max_beauty(s: String, k: i32) -> i32 {\n\n    }\n}",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Racket",
                    "langSlug": "racket",
                    "code": "(define/contract (count-k-subsequences-with-max-beauty s k)\n  (-> string? exact-integer? exact-integer?)\n\n  )",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Erlang",
                    "langSlug": "erlang",
                    "code": "-spec count_k_subsequences_with_max_beauty(S :: unicode:unicode_binary(), K :: integer()) -> integer().\ncount_k_subsequences_with_max_beauty(S, K) ->\n  .",
                    "__typename": "CodeSnippetNode"
                },
                {
                    "lang": "Elixir",
                    "langSlug": "elixir",
                    "code": "defmodule Solution do\n  @spec count_k_subsequences_with_max_beauty(s :: String.t, k :: integer) :: integer\n  def count_k_subsequences_with_max_beauty(s, k) do\n\n  end\nend",
                    "__typename": "CodeSnippetNode"
                }
            ],
            "stats": "{\"totalAccepted\": \"2.6K\", \"totalSubmission\": \"9.1K\", \"totalAcceptedRaw\": 2584, \"totalSubmissionRaw\": 9062, \"acRate\": \"28.5%\"}",
            "hints": [
                "Since every character appears once in a k-subsequence, we can solve the following problem first: Find the total number of ways to select <code>k</code> characters such that the sum of their frequencies is maximum.",
                "An obvious case to eliminate is if <code>k</code> is greater than the number of distinct characters in <code>s</code>, then the answer is <code>0</code>.",
                "We are now interested in the top frequencies among the characters. Using a map data structure, let <code>cnt[x]</code> denote the number of characters that have a frequency of <code>x</code>.",
                "Starting from the maximum value <code>x</code> in <code>cnt</code>. Let <code>i = min(k, cnt[x])</code> we add to our result <code> <sup>cnt[x]</sup>C<sub>i</sub> * x<sup>i</sup></code> representing the number of ways to select <code>i</code> characters from all characters with frequency <code>x</code>, multiplied by the number of ways to choose each individual character. Subtract <code>i</code> from <code>k</code> and continue downwards to the next maximum value.",
                "Powers, combinations, and additions should be done modulo <code>10<sup>9</sup> + 7</code>."
            ],
            "solution": null,
            "status": null,
            "sampleTestCase": "\"bcca\"\n2",
            "metaData": "{\n  \"name\": \"countKSubsequencesWithMaxBeauty\",\n  \"params\": [\n    {\n      \"name\": \"s\",\n      \"type\": \"string\"\n    },\n    {\n      \"type\": \"integer\",\n      \"name\": \"k\"\n    }\n  ],\n  \"return\": {\n    \"type\": \"integer\"\n  }\n}",
            "judgerAvailable": true,
            "judgeType": "large",
            "mysqlSchemas": [],
            "enableRunCode": true,
            "envInfo": "{\"cpp\":[\"C++\",\"<p>\\u7248\\u672c\\uff1a<code>clang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 20\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528<code>-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002<a href=\\\"https:\\/\\/github.com\\/google\\/sanitizers\\/wiki\\/AddressSanitizer\\\" target=\\\"_blank\\\">AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4b<code>out-of-bounds<\\/code>\\u548c<code>use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\"<p>\\u7248\\u672c\\uff1a<code>OpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\"<p>\\u7248\\u672c\\uff1a <code>Python 2.7.12<\\/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\\uff1a<a href=\\\"https:\\/\\/docs.python.org\\/2\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/2\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/2\\/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>\\u6ce8\\u610f Python 2.7 <a href=\\\"https:\\/\\/www.python.org\\/dev\\/peps\\/pep-0373\\/\\\" target=\\\"_blank\\\">\\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\"<p>\\u7248\\u672c\\uff1a<code>GCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528<code>-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 <a href=\\\"https:\\/\\/github.com\\/google\\/sanitizers\\/wiki\\/AddressSanitizer\\\" target=\\\"_blank\\\">AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4b<code>out-of-bounds<\\/code>\\u548c<code>use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n    HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n<p><b>2. \\u5728\\u54c8\\u5e0c\\u8868\\u4e2d\\u67e5\\u627e\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n    struct hash_entry *s;\\r\\n    HASH_FIND_INT(users, &user_id, s);\\r\\n    return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n<p><b>3. \\u4ece\\u54c8\\u5e0c\\u8868\\u4e2d\\u5220\\u9664\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n    HASH_DEL(users, user);  \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"<p><a href=\\\"https:\\/\\/docs.microsoft.com\\/en-us\\/dotnet\\/csharp\\/whats-new\\/csharp-9\\\" target=\\\"_blank\\\">C# 10<\\/a> \\u8fd0\\u884c\\u5728 .NET 6 \\u4e0a<\\/p>\"],\"javascript\":[\"JavaScript\",\"<p>\\u7248\\u672c\\uff1a<code>Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5728\\u6267\\u884c\\u65f6\\u5c06\\u5e26\\u4e0a <code>--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f <a href=\\\"http:\\/\\/node.green\\/\\\" target=\\\"_blank\\\">\\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/tree\\/fb4fdb984834421279aeb081df7af624d17c2a03\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/e63563025a5a805aa16928cb53bcd517bfea9230\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.1<\\/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 1.21<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 <a href=\\\"https:\\/\\/github.com\\/emirpasic\\/gods\\/tree\\/v1.18.1\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods@v1.18.1<\\/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 1.9.0<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u4f7f\\u7528\\u7684\\u662f JetBrains \\u63d0\\u4f9b\\u7684 experimental compiler\\u3002\\u5982\\u679c\\u60a8\\u8ba4\\u4e3a\\u60a8\\u9047\\u5230\\u4e86\\u7f16\\u8bd1\\u5668\\u76f8\\u5173\\u7684\\u95ee\\u9898\\uff0c\\u8bf7\\u5411\\u6211\\u4eec\\u53cd\\u9988<\\/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 5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/tree\\/fb4fdb984834421279aeb081df7af624d17c2a03\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.3.0<\\/a> \\u548c <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/e63563025a5a805aa16928cb53bcd517bfea9230\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.1<\\/a>\\u3002<\\/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\"],\"dart\":[\"Dart\",\"<p>Dart 2.17.3<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"]}",
            "book": null,
            "isSubscribed": false,
            "isDailyQuestion": false,
            "dailyRecordStatus": null,
            "editorType": "CKEDITOR",
            "ugcQuestionId": null,
            "style": "LEETCODE",
            "exampleTestcases": "\"bcca\"\n2\n\"abbcd\"\n4",
            "__typename": "QuestionNode"
        }
    }
}