2452. Words Within Two Edits of Dictionary

Explanation:

To solve this problem, we can iterate through each word in the queries array and compare it with each word in the dictionary array. For each pair of words, we check if they are equal or if they can be made equal with at most two edits. We can achieve this by comparing the characters in both words and keeping track of the differences.

For each word in queries, we iterate through each word in dictionary.
If the words are equal, or if they can be made equal with at most two edits, we add the word from queries to the result list.
We return the list of words from queries that match with some word from dictionary after a maximum of two edits.

Time complexity: O(n * m * k), where n is the number of words in queries, m is the number of words in dictionary, and k is the length of each word. Space complexity: O(n), where n is the number of words in queries.

:

View Leetcode Problem

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> wordMatches(String[] queries, String[] dictionary) {
        List<String> result = new ArrayList<>();
        
        for (String query : queries) {
            for (String word : dictionary) {
                if (query.equals(word) || areTwoEditsApart(query, word)) {
                    result.add(query);
                    break;
                }
            }
        }
        
        return result;
    }
    
    private boolean areTwoEditsApart(String word1, String word2) {
        int diffCount = 0;
        
        for (int i = 0; i < word1.length(); i++) {
            if (word1.charAt(i) != word2.charAt(i)) {
                diffCount++;
            }
            if (diffCount > 2) {
                return false;
            }
        }
        
        return true;
    }
}

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Explanation:

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