You are given two string arrays, queries and dictionary. All words in each array comprise of lowercase English letters and have the same length.
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.
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.
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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\n1 <= queries.length, dictionary.length <= 100n == queries[i].length == dictionary[j].length1 <= n <= 100queries[i] and dictionary[j] are composed of lowercase English letters.Compiled with clang 11 using the latest C++ 17 standard.
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