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:
- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".
- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".
- It would take more than 2 edits for "ants" to equal a dictionary word.
- "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] and dictionary[j] are composed of lowercase English letters.

Solution: Hamming distance + Brute Force

For each query word q, check the hamming distance between it and all words in the dictionary.

Time complexity: O(|q|*|d|*n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  vector<string> twoEditWords(vector<string>& queries, vector<string>& dictionary) {
    const int n = queries[0].size();
    auto check = [&](string_view q) {      
      for (const string& w : dictionary) {
        int dist = 0;
        for (int i = 0; i < n && dist <= 3; ++i)
          dist += q[i] != w[i];
        if (dist <= 2) return true;
      }
      return false;
    };
    vector<string> ans;
    for (const string& q : queries) 
      if (check(q)) ans.push_back(q);
    return ans;
  }
};

The post 花花酱 LeetCode 2452. Words Within Two Edits of Dictionary appeared first on Huahua's Tech Road.