Class Solution
Medium
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
.
Example 1:
Input: queries = [“word”,“note”,“ants”,“wood”], dictionary = [“wood”,“joke”,“moat”]
Output: [“word”,“note”,“wood”]
Explanation:
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Changing the ‘r’ in “word” to ‘o’ allows it to equal the dictionary word “wood”.
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Changing the ‘n’ to ‘j’ and the ‘t’ to ‘k’ in “note” changes it to “joke”.
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It would take more than 2 edits for “ants” to equal a dictionary word.
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“wood” can remain unchanged (0 edits) and match the corresponding dictionary word.
Thus, we return [“word”,“note”,“wood”].
Example 2:
Input: queries = [“yes”], dictionary = [“not”]
Output: []
Explanation:
Applying any two edits to “yes” cannot make it equal to “not”. Thus, we return an empty array.
Constraints:
1 <= queries.length, dictionary.length <= 100
n == queries[i].length == dictionary[j].length
1 <= n <= 100
- All
queries[i]
anddictionary[j]
are composed of lowercase English letters.
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Constructor Summary
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Method Summary
Modifier and TypeMethodDescriptionjava.util.List
<java.lang.String> twoEditWords
(java.lang.String[] queries, java.lang.String[] dictionary) Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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twoEditWords
public java.util.List<java.lang.String> twoEditWords(java.lang.String[] queries, java.lang.String[] dictionary)
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