Introduction

In this blog post, we will explore how to solve the LeetCode problem "2452" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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: - 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.

Explanation

• Iterate and Compare: Iterate through each query word and compare it with every word in the dictionary.
  • Edit Distance Calculation: For each query-dictionary word pair, calculate the edit distance (number of differing characters).
  • Filter and Collect: If the edit distance is less than or equal to 2, add the query word to the result list.

• Time Complexity: O(Q D N), where Q is the number of queries, D is the number of words in the dictionary, and N is the length of each word. Space Complexity: O(Q), where Q is the number of queries (for the result list).

def solve():
    queries = ["word","note","ants","wood"]
    dictionary = ["wood","joke","moat"]
    expected = ["word","note","wood"]
    assert find_words(queries, dictionary) == expected

    queries = ["yes"]
    dictionary = ["not"]
    expected = []
    assert find_words(queries, dictionary) == expected

def find_words(queries, dictionary):
    result = []
    for query in queries:
        for word in dictionary:
            diff_count = 0
            for i in range(len(query)):
                if query[i] != word[i]:
                    diff_count += 1
            if diff_count <= 2:
                result.append(query)
                break  # Move to the next query word if a match is found
    return result


    # Code
    ```python
    def find_words(queries, ditionary):    result = []
    for query in queries:
        for word in dictionary:
            diff_count = 0
            for i in range(len(query)):
                if query[i] != word[i]:
                    diff_count += 1
            if diff_count <= 2:
                result.append(query)
                break
    return result

Summary

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