{ "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, queries and dictionary. All words in each array comprise of lowercase English letters and have the same length.

\n\n

In one edit you can take a word from 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.

\n\n

Return a list of all words from 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.

\n\n

 

\n

Example 1:

\n\n
\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

Example 2:

\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

Constraints:

\n\n\n", "translatedTitle": "距离字典两次编辑以内的单词", "translatedContent": "

给你两个字符串数组 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。

\n\n

一次 编辑 中,你可以从 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。

\n\n

请你返回 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。

\n\n

 

\n\n

示例 1:

\n\n
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

示例 2:

\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
\n\n

 

\n\n

提示:

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