2452. Words Within Two Edits of Dictionary

/**
 * Problem: LeetCode 2452 - Words Within Two Edits of Dictionary
 * Difficulty: Medium
 * Language: Java
 * Author: CodeByTushu
 */
import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> twoEditWords(String[] queries, String[] dictionary) {
        List<String> result = new ArrayList<>();
        
        for (String query : queries) {
            for (String dictWord : dictionary) {
                if (canMatchWithTwoEdits(query, dictWord)) {
                    result.add(query);
                    break;
                }
            }
        }
        
        return result;
    }
    
    private boolean canMatchWithTwoEdits(String s1, String s2) {
        int diff = 0;
        int len = s1.length();
        for (int i = 0; i < len; i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                diff++;
                if (diff > 2) return false;
            }
        }
        return true;
    }
}

"""
Problem: LeetCode 2452 - Words Within Two Edits of Dictionary
Difficulty: Medium
Language: Python
Author: CodeByTushu
"""
from typing import List


class Solution:
    def twoEditWords(
        self, queries: List[str], dictionary: List[str]
    ) -> List[str]:
        result = []

        for query in queries:
            for word in dictionary:
                diff = 0
                for c1, c2 in zip(query, word):
                    if c1 != c2:
                        diff += 1
                        if diff > 2:
                            break
                if diff <= 2:
                    result.append(query)
                    break

        return result

/**
 * Problem: LeetCode 2452 - Words Within Two Edits of Dictionary
 * Difficulty: Medium
 * Language: C++
 * Author: CodeByTushu
 */
#include <vector>
#include <string>

using namespace std;

class Solution {
public:
    vector<string> twoEditWords(vector<string>& queries, vector<string>& dictionary) {
        vector<string> result;
        int wordLen = queries[0].length();
        
        for (const string& query : queries) {
            for (const string& word : dictionary) {
                int diff = 0;
                for (int i = 0; i < wordLen; ++i) {
                    if (query[i] != word[i]) {
                        if (++diff > 2) break;
                    }
                }
                if (diff <= 2) {
                    result.push_back(query);
                    break;
                }
            }
        }
        
        return result;
    }
};

/**
 * Problem: LeetCode 2452 - Words Within Two Edits of Dictionary
 * Difficulty: Medium
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {string[]} queries
 * @param {string[]} dictionary
 * @return {string[]}
 */
var twoEditWords = function(queries, dictionary) {
    const result = [];
    const wordLen = queries[0].length;

    for (let q = 0; q < queries.length; q++) {
        const query = queries[q];

        for (let d = 0; d < dictionary.length; d++) {
            const word = dictionary[d];
            let diff = 0;

            for (let i = 0; i < wordLen; i++) {
                if (query.charCodeAt(i) !== word.charCodeAt(i)) {
                    diff++;
                    if (diff > 2) break;
                }
            }

            if (diff <= 2) {
                result.push(query);
                break;
            }
        }
    }

    return result;
};