n.",
+ "If there is a prime factor p such that p >= 11, the answer is -1. Since there are no digits whose products equal p.",
+ "Factors 5 and 7 should be included in the answer since their product with any number bigger than 1 is a 2-digit number.",
+ "For factors 2 and 3, we group every three 2 into an 8 and every two 3 into a 9.",
+ "For any leftover 2 or 3, check all the possible combinations."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "105",
+ "metaData": "{\n \"name\": \"smallestNumber\",\n \"params\": [\n {\n \"name\": \"n\",\n \"type\": \"long\"\n }\n ],\n \"return\": {\n \"type\": \"string\"\n }\n}",
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diff --git a/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-i.json b/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-i.json
new file mode 100644
index 00000000..9ddee295
--- /dev/null
+++ b/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-i.json
@@ -0,0 +1,171 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2999",
+ "questionFrontendId": "2839",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2420791,
+ "title": "Check if Strings Can be Made Equal With Operations I",
+ "titleSlug": "check-if-strings-can-be-made-equal-with-operations-i",
+ "content": " You are given two strings You can apply the following operation on any of the two strings any number of times: Return Example 1: Example 2: Constraints: 给你两个字符串 你可以对两个字符串中的 任意一个 执行以下操作 任意 次： 如果你可以让字符串 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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\",\" \\u7248\\u672c\\uff1a \\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 \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a 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diff --git a/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-ii.json b/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-ii.json
new file mode 100644
index 00000000..a4c44d86
--- /dev/null
+++ b/leetcode-cn/originData/check-if-strings-can-be-made-equal-with-operations-ii.json
@@ -0,0 +1,183 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2978",
+ "questionFrontendId": "2840",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2420797,
+ "title": "Check if Strings Can be Made Equal With Operations II",
+ "titleSlug": "check-if-strings-can-be-made-equal-with-operations-ii",
+ "content": " You are given two strings You can apply the following operation on any of the two strings any number of times: Return Example 1: Example 2: Constraints: 给你两个字符串 你可以对两个字符串中的 任意一个 执行以下操作 任意 次： 如果你可以让字符串 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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\",\" \\u7248\\u672c\\uff1a \\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 \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a 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\ No newline at end of file
diff --git a/leetcode-cn/originData/count-k-subsequences-of-a-string-with-maximum-beauty.json b/leetcode-cn/originData/count-k-subsequences-of-a-string-with-maximum-beauty.json
new file mode 100644
index 00000000..ac888b29
--- /dev/null
+++ b/leetcode-cn/originData/count-k-subsequences-of-a-string-with-maximum-beauty.json
@@ -0,0 +1,198 @@
+{
+ "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": " You are given a string A k-subsequence is a subsequence of Let The beauty of a k-subsequence is the sum of For example, consider Return an integer denoting the number of k-subsequences whose beauty is the maximum among all k-subsequences. Since the answer may be too large, return it modulo 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. Notes Example 1: Example 2: Constraints: 给你一个字符串 k 子序列指的是 定义 k 子序列的 美丽值 定义为这个子序列中每一个字符 比方说， 请你返回一个整数，表示所有 k 子序列 里面 美丽值 是 最大值 的子序列数目。由于答案可能很大，将结果对 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。 注意： 示例 1： 示例 2： 提示： \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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\",\" \\u7248\\u672c\\uff1a \\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 \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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 \\u6ce8\\u610f Python 2.7 \\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\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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 \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\nclang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 20\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU11\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\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\\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\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nRuby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\nSwift 5.5.2<\\/code><\\/p>\\r\\n\\r\\nGo 1.18<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\nrust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\ns1 and s2, both of length 4, consisting of lowercase English letters.\n\t
\n\ni and j such that j - i = 2, then swap the two characters at those indices in the string.true if you can make the strings s1 and s2 equal, and false otherwise.\nInput: s1 = "abcd", s2 = "cdab"\nOutput: true\nExplanation: We can do the following operations on s1:\n- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbad".\n- Choose the indices i = 1, j = 3. The resulting string is s1 = "cdab" = s2.\n
\n\n\nInput: s1 = "abcd", s2 = "dacb"\nOutput: false\nExplanation: It is not possible to make the two strings equal.\n
\n\n\n\t
\n",
+ "translatedTitle": "判断通过操作能否让字符串相等 I",
+ "translatedContent": "s1.length == s2.length == 4s1 and s2 consist only of lowercase English letters.s1 和 s2 ，两个字符串的长度都为 4 ，且只包含 小写 英文字母。\n\t
\n\ni 和 j 且满足 j - i = 2 ，然后 交换 这个字符串中两个下标对应的字符。s1 和 s2 相等，那么返回 true ，否则返回 false 。\n输入：s1 = \"abcd\", s2 = \"cdab\"\n输出：true\n解释： 我们可以对 s1 执行以下操作：\n- 选择下标 i = 0 ，j = 2 ，得到字符串 s1 = \"cbad\" 。\n- 选择下标 i = 1 ，j = 3 ，得到字符串 s1 = \"cdab\" = s2 。\n
\n\n\n输入：s1 = \"abcd\", s2 = \"dacb\"\n输出：false\n解释：无法让两个字符串相等。\n
\n\n\n\t
\n",
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\n\n\nInput: s1 = "abe", s2 = "bea"\nOutput: false\nExplanation: It is not possible to make the two strings equal.\n
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+ "translatedTitle": "判断通过操作能否让字符串相等 II",
+ "translatedContent": "n == s1.length == s2.length1 <= n <= 105s1 and s2 consist only of lowercase English letters.s1 和 s2 ，两个字符串长度都为 n ，且只包含 小写 英文字母。\n\t
\n\ni 和 j ，满足 i < j 且 j - i 是 偶数，然后 交换 这个字符串中两个下标对应的字符。s1 和 s2 相等，那么返回 true ，否则返回 false 。\n输入：s1 = \"abcdba\", s2 = \"cabdab\"\n输出：true\n解释：我们可以对 s1 执行以下操作：\n- 选择下标 i = 0 ，j = 2 ，得到字符串 s1 = \"cbadba\" 。\n- 选择下标 i = 2 ，j = 4 ，得到字符串 s1 = \"cbbdaa\" 。\n- 选择下标 i = 1 ，j = 5 ，得到字符串 s1 = \"cabdab\" = s2 。\n
\n\n\n输入：s1 = \"abe\", s2 = \"bea\"\n输出：false\n解释：无法让两个字符串相等。\n
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\n\t"abbbdd" -> "ab" having a beauty of f('a') + f('b') = 4"abbbdd" -> "ad" having a beauty of f('a') + f('d') = 3"abbbdd" -> "bd" having a beauty of f('b') + f('d') = 5109 + 7.\n\t
\n\nf(c) is the number of times a character c occurs in s, not a k-subsequence.\nInput: s = "bcca", k = 2\nOutput: 4\nExplanation: From s we have f('a') = 1, f('b') = 1, and f('c') = 2.\nThe k-subsequences of s are: \nbcca having a beauty of f('b') + f('c') = 3 \nbcca having a beauty of f('b') + f('c') = 3 \nbcca having a beauty of f('b') + f('a') = 2 \nbcca having a beauty of f('c') + f('a') = 3\nbcca having a beauty of f('c') + f('a') = 3 \nThere are 4 k-subsequences that have the maximum beauty, 3. \nHence, the answer is 4. \n\n\n\nInput: s = "abbcd", k = 4\nOutput: 2\nExplanation: From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. \nThe k-subsequences of s are: \nabbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5\nabbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5 \nThere are 2 k-subsequences that have the maximum beauty, 5. \nHence, the answer is 2. \n\n\n\n\t
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+ "translatedTitle": "统计一个字符串的 k 子序列美丽值最大的数目",
+ "translatedContent": "1 <= s.length <= 2 * 1051 <= k <= s.lengths consists only of lowercase English letters.s 和一个整数 k 。s 的一个长度为 k 的 子序列 ，且所有字符都是 唯一 的，也就是说每个字符在子序列里只出现过一次。f(c) 为字符 c 在 s 中出现的次数。c 的 f(c) 之 和 。s = \"abbbdd\" 和 k = 2 ，我们有：\n\t
\n\nf('a') = 1, f('b') = 3, f('d') = 2s 的部分 k 子序列为：\n\t\n\t\t
\n\t\"abbbdd\" -> \"ab\" ，美丽值为 f('a') + f('b') = 4\"abbbdd\" -> \"ad\" ，美丽值为 f('a') + f('d') = 3\"abbbdd\" -> \"bd\" ，美丽值为 f('b') + f('d') = 5109 + 7 取余后返回。\n\t
\n\nf(c) 指的是字符 c 在字符串 s 的出现次数，不是在 k 子序列里的出现次数。\n输入：s = \"bcca\", k = 2\n输出：4\n解释：s 中我们有 f('a') = 1 ，f('b') = 1 和 f('c') = 2 。\ns 的 k 子序列为：\nbcca ，美丽值为 f('b') + f('c') = 3\nbcca ，美丽值为 f('b') + f('c') = 3\nbcca ，美丽值为 f('b') + f('a') = 2\nbcca ，美丽值为 f('c') + f('a') = 3\nbcca ，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\n\n\n输入：s = \"abbcd\", k = 4\n输出：2\n解释：s 中我们有 f('a') = 1 ，f('b') = 2 ，f('c') = 1 和 f('d') = 1 。\ns 的 k 子序列为：\nabbcd ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5\nabbcd ，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
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+ "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\": \"2.2K\", \"totalSubmission\": \"8K\", \"totalAcceptedRaw\": 2186, \"totalSubmissionRaw\": 7999, \"acRate\": \"27.3%\"}",
+ "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 1 <= s.length <= 2 * 1051 <= k <= s.lengths 只包含小写英文字母。k characters such that the sum of their frequencies is maximum.",
+ "An obvious case to eliminate is if k is greater than the number of distinct characters in s, then the answer is 0.",
+ "We are now interested in the top frequencies among the characters. Using a map data structure, let cnt[x] denote the number of characters that have a frequency of x.",
+ "Starting from the maximum value x in cnt. Let i = min(k, cnt[x]) we add to our result cnt[x]Ci * xi representing the number of ways to select i characters from all characters with frequency x, multiplied by the number of ways to choose each individual character. Subtract i from k and continue downwards to the next maximum value.",
+ "Powers, combinations, and additions should be done modulo 109 + 7."
+ ],
+ "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}",
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