You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.
The second greater integer of nums[i] is nums[j] such that:
j > inums[j] > nums[i]k such that nums[k] > nums[i] and i < k < j.If there is no such nums[j], the second greater integer is considered to be -1.
[1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1.Return an integer array answer, where answer[i] is the second greater integer of nums[i].
\n
Example 1:
\n\n\nInput: nums = [2,4,0,9,6]\nOutput: [9,6,6,-1,-1]\nExplanation:\n0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.\n1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.\n2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.\n3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.\n4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.\nThus, we return [9,6,6,-1,-1].\n\n\n
Example 2:
\n\n\nInput: nums = [3,3]\nOutput: [-1,-1]\nExplanation:\nWe return [-1,-1] since neither integer has any integer greater than it.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1050 <= nums[i] <= 109给你一个下标从 0 开始的非负整数数组 nums 。对于 nums 中每一个整数，你必须找到对应元素的 第二大 整数。
如果 nums[j] 满足以下条件，那么我们称它为 nums[i] 的 第二大 整数：
j > inums[j] > nums[i]k 满足 i < k < j 且 nums[k] > nums[i] 。如果不存在 nums[j] ，那么第二大整数为 -1 。
[1, 2, 4, 3] 中，1 的第二大整数是 4 ，2 的第二大整数是 3 ，3 和 4 的第二大整数是 -1 。请你返回一个整数数组 answer ，其中 answer[i]是 nums[i] 的第二大整数。
\n\n
示例 1：
\n\n\n输入：nums = [2,4,0,9,6]\n输出：[9,6,6,-1,-1]\n解释：\n下标为 0 处：2 的右边，4 是大于 2 的第一个整数，9 是第二个大于 2 的整数。\n下标为 1 处：4 的右边，9 是大于 4 的第一个整数，6 是第二个大于 4 的整数。\n下标为 2 处：0 的右边，9 是大于 0 的第一个整数，6 是第二个大于 0 的整数。\n下标为 3 处：右边不存在大于 9 的整数，所以第二大整数为 -1 。\n下标为 4 处：右边不存在大于 6 的整数，所以第二大整数为 -1 。\n所以我们返回 [9,6,6,-1,-1] 。\n\n\n
示例 2：
\n\n\n输入：nums = [3,3]\n输出：[-1,-1]\n解释：\n由于每个数右边都没有更大的数，所以我们返回 [-1,-1] 。\n\n\n
\n\n
提示：
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\ No newline at end of file
diff --git a/leetcode-cn/originData/words-within-two-edits-of-dictionary.json b/leetcode-cn/originData/words-within-two-edits-of-dictionary.json
new file mode 100644
index 00000000..543be147
--- /dev/null
+++ b/leetcode-cn/originData/words-within-two-edits-of-dictionary.json
@@ -0,0 +1,165 @@
+{
+ "data": {
+ "question": {
+ "questionId": "2550",
+ "questionFrontendId": "2452",
+ "categoryTitle": "Algorithms",
+ "boundTopicId": 1930603,
+ "title": "Words Within Two Edits of Dictionary",
+ "titleSlug": "words-within-two-edits-of-dictionary",
+ "content": " You are given two string arrays, In one edit you can take a word from Return a list of all words from Example 1: Example 2: Constraints: 给你两个字符串数组 一次 编辑 中，你可以从 请你返回 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 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8.1<\\/code>.<\\/p>\\r\\n\\r\\nqueries and dictionary. All words in each array comprise of lowercase English letters and have the same length.queries, and change any letter in it to any other letter. Find all words from queries that, after a maximum of two edits, equal some word from dictionary.queries, that match with some word from dictionary after a maximum of two edits. Return the words in the same order they appear in queries.\nInput: queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]\nOutput: ["word","note","wood"]\nExplanation:\n- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".\n- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".\n- It would take more than 2 edits for "ants" to equal a dictionary word.\n- "wood" can remain unchanged (0 edits) and match the corresponding dictionary word.\nThus, we return ["word","note","wood"].\n
\n\n\nInput: queries = ["yes"], dictionary = ["not"]\nOutput: []\nExplanation:\nApplying any two edits to "yes" cannot make it equal to "not". Thus, we return an empty array.\n
\n\n\n\t
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+ "translatedTitle": "距离字典两次编辑以内的单词",
+ "translatedContent": "1 <= queries.length, dictionary.length <= 100n == queries[i].length == dictionary[j].length1 <= n <= 100queries[i] and dictionary[j] are composed of lowercase English letters.queries 和 dictionary 。数组中所有单词都只包含小写英文字母，且长度都相同。queries 中选择一个单词，将任意一个字母修改成任何其他字母。从 queries 中找到所有满足以下条件的字符串：不超过 两次编辑内，字符串与 dictionary 中某个字符串相同。queries 中的单词列表，这些单词距离 dictionary 中的单词 编辑次数 不超过 两次 。单词返回的顺序需要与 queries 中原本顺序相同。输入：queries = [\"word\",\"note\",\"ants\",\"wood\"], dictionary = [\"wood\",\"joke\",\"moat\"]\n输出：[\"word\",\"note\",\"wood\"]\n解释：\n- 将 \"word\" 中的 'r' 换成 'o' ，得到 dictionary 中的单词 \"wood\" 。\n- 将 \"note\" 中的 'n' 换成 'j' 且将 't' 换成 'k' ，得到 \"joke\" 。\n- \"ants\" 需要超过 2 次编辑才能得到 dictionary 中的单词。\n- \"wood\" 不需要修改（0 次编辑），就得到 dictionary 中相同的单词。\n所以我们返回 [\"word\",\"note\",\"wood\"] 。\n
\n\n输入：queries = [\"yes\"], dictionary = [\"not\"]\n输出：[]\n解释：\n\"yes\" 需要超过 2 次编辑才能得到 \"not\" 。\n所以我们返回空数组。\n
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+ "code": "class Solution {\npublic:\n vector1 <= queries.length, dictionary.length <= 100n == queries[i].length == dictionary[j].length1 <= n <= 100queries[i] 和 dictionary[j] 都只包含小写英文字母。clang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 17\\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 1.17<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\nScala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\nPHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n