O(1) using sparse tables and preprocessing.",
+ "For every index L, let’s find the subarray starting at the index L and maximizing gcd-sum.",
+ "Use the fact that if L is fixed, then by adding one more element to the end of a subarray, two things can happen: the gcd remains the same as the last gcd or becomes at least half of the last one.",
+ "Now we can use binary search to find the last index R such that gcd of the elements of nums[L..R] would be equal to nums[L].",
+ "Now add nums[R + 1] to the current subarray and continue the process to find the last index that has the same gcd as the current gcd of elements."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "[2,1,4,4,4,2]\n2",
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\ No newline at end of file
diff --git a/leetcode-cn/originData/[no content]number-of-divisible-substrings.json b/leetcode-cn/originData/[no content]number-of-divisible-substrings.json
new file mode 100644
index 00000000..ca4dc406
--- /dev/null
+++ b/leetcode-cn/originData/[no content]number-of-divisible-substrings.json
@@ -0,0 +1,69 @@
+{
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+ "questionId": "3237",
+ "questionFrontendId": "2950",
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+ "boundTopicId": 2547518,
+ "title": "Number of Divisible Substrings",
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+ {
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+ "translatedName": "字符串",
+ "__typename": "TopicTagNode"
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+ "Iterate over all substrings in \\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>, 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\ No newline at end of file
diff --git a/leetcode-cn/originData/[no content]number-of-same-end-substrings.json b/leetcode-cn/originData/[no content]number-of-same-end-substrings.json
new file mode 100644
index 00000000..c8a43a3b
--- /dev/null
+++ b/leetcode-cn/originData/[no content]number-of-same-end-substrings.json
@@ -0,0 +1,49 @@
+{
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+ "question": {
+ "questionId": "3247",
+ "questionFrontendId": "2955",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2557105,
+ "title": "Number of Same-End Substrings",
+ "titleSlug": "number-of-same-end-substrings",
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+ "stats": "{\"totalAccepted\": \"79\", \"totalSubmission\": \"83\", \"totalAcceptedRaw\": 79, \"totalSubmissionRaw\": 83, \"acRate\": \"95.2%\"}",
+ "hints": [
+ "If there are \\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>, 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+ "book": null,
+ "isSubscribed": false,
+ "isDailyQuestion": false,
+ "dailyRecordStatus": null,
+ "editorType": "CKEDITOR",
+ "ugcQuestionId": null,
+ "style": "LEETCODE",
+ "exampleTestcases": "\"abcaab\"\n[[0,0],[1,4],[2,5],[0,5]]\n\"abcd\"\n[[0,3]]",
+ "__typename": "QuestionNode"
+ }
+ }
+}
\ No newline at end of file
diff --git a/leetcode-cn/originData/count-beautiful-substrings-i.json b/leetcode-cn/originData/count-beautiful-substrings-i.json
new file mode 100644
index 00000000..0a2a2c9d
--- /dev/null
+++ b/leetcode-cn/originData/count-beautiful-substrings-i.json
@@ -0,0 +1,182 @@
+{
+ "data": {
+ "question": {
+ "questionId": "3210",
+ "questionFrontendId": "2947",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 2538343,
+ "title": "Count Beautiful Substrings I",
+ "titleSlug": "count-beautiful-substrings-i",
+ "content": " You are given a string Let A string is beautiful if: Return the number of non-empty beautiful substrings in the given string A substring is a contiguous sequence of characters in a string. Vowel letters in English are Consonant letters in English are every letter except vowels. Example 1: Example 2: Example 3: Constraints: 给你一个字符串 用 如果某个字符串满足以下条件，则称其为 美丽字符串 ： 返回字符串 子字符串是字符串中的一个连续字符序列。 英语中的 元音字母 为 英语中的 辅音字母 为除了元音字母之外的所有字母。 示例 1： 示例 2： 示例 3： 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 1.21<\\/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\\nO(n * n).",
+ "For each substring, try to calculate the sum of the mapped values in O(1).",
+ "To do the above, use a partial sum array."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "\"asdf\"",
+ "metaData": "{\n \"name\": \"countDivisibleSubstrings\",\n \"params\": [\n {\n \"name\": \"word\",\n \"type\": \"string\"\n }\n ],\n \"return\": {\n \"type\": \"integer\"\n }\n}",
+ "judgerAvailable": true,
+ "judgeType": "large",
+ "mysqlSchemas": [],
+ "enableRunCode": true,
+ "envInfo": "{\"cpp\":[\"C++\",\"clang 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 1.21<\\/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\\nt occurrences of a character in a substring, there exists t * (t - 1) / 2 Same-End substrings with that character.",
+ "Try to calculate the number of occurrences of a character in a substring in O(1) using partial sum."
+ ],
+ "solution": null,
+ "status": null,
+ "sampleTestCase": "\"abcaab\"\n[[0,0],[1,4],[2,5],[0,5]]",
+ "metaData": "{\n \"name\": \"sameEndSubstringCount\",\n \"params\": [\n {\n \"name\": \"s\",\n \"type\": \"string\"\n },\n {\n \"type\": \"integer[][]\",\n \"name\": \"queries\"\n }\n ],\n \"return\": {\n \"type\": \"integer[]\"\n }\n}",
+ "judgerAvailable": true,
+ "judgeType": "large",
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\n\nvowels == consonants.(vowels * consonants) % k == 0, in other terms the multiplication of vowels and consonants is divisible by k.s.'a', 'e', 'i', 'o', and 'u'.\nInput: s = "baeyh", k = 2\nOutput: 2\nExplanation: There are 2 beautiful substrings in the given string.\n- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["y","h"]).\nYou can see that string "aeyh" is beautiful as vowels == consonants and vowels * consonants % k == 0.\n- Substring "baeyh", vowels = 2 (["a",e"]), consonants = 2 (["b","y"]). \nYou can see that string "baey" is beautiful as vowels == consonants and vowels * consonants % k == 0.\nIt can be shown that there are only 2 beautiful substrings in the given string.\n
\n\n\nInput: s = "abba", k = 1\nOutput: 3\nExplanation: There are 3 beautiful substrings in the given string.\n- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]). \n- Substring "abba", vowels = 1 (["a"]), consonants = 1 (["b"]).\n- Substring "abba", vowels = 2 (["a","a"]), consonants = 2 (["b","b"]).\nIt can be shown that there are only 3 beautiful substrings in the given string.\n
\n\n\nInput: s = "bcdf", k = 1\nOutput: 0\nExplanation: There are no beautiful substrings in the given string.\n
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+ "translatedTitle": "统计美丽子字符串 I",
+ "translatedContent": "1 <= s.length <= 10001 <= k <= 1000s consists of only English lowercase letters.s 和一个正整数 k 。vowels 和 consonants 分别表示字符串中元音字母和辅音字母的数量。\n\t
\n\nvowels == consonants，即元音字母和辅音字母的数量相等。(vowels * consonants) % k == 0，即元音字母和辅音字母的数量的乘积能被 k 整除。s 中 非空美丽子字符串 的数量。'a'、'e'、'i'、'o' 和 'u' 。\n输入：s = \"baeyh\", k = 2\n输出：2\n解释：字符串 s 中有 2 个美丽子字符串。\n- 子字符串 \"baeyh\"，vowels = 2（[\"a\",\"e\"]），consonants = 2（[\"y\",\"h\"]）。\n可以看出字符串 \"aeyh\" 是美丽字符串，因为 vowels == consonants 且 vowels * consonants % k == 0 。\n- 子字符串 \"baeyh\"，vowels = 2（[\"a\",\"e\"]），consonants = 2（[\"b\",\"y\"]）。\n可以看出字符串 \"baey\" 是美丽字符串，因为 vowels == consonants 且 vowels * consonants % k == 0 。\n可以证明字符串 s 中只有 2 个美丽子字符串。\n
\n\n\n输入：s = \"abba\", k = 1\n输出：3\n解释：字符串 s 中有 3 个美丽子字符串。\n- 子字符串 \"abba\"，vowels = 1（[\"a\"]），consonants = 1（[\"b\"]）。\n- 子字符串 \"abba\"，vowels = 1（[\"a\"]），consonants = 1（[\"b\"]）。\n- 子字符串 \"abba\"，vowels = 2（[\"a\",\"a\"]），consonants = 2（[\"b\",\"b\"]）。\n可以证明字符串 s 中只有 3 个美丽子字符串。\n
\n\n\n输入：s = \"bcdf\", k = 1\n输出：0\n解释：字符串 s 中没有美丽子字符串。\n
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