Introduction

In this blog post, we will explore how to solve the LeetCode problem "676" 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

Design a data structure that is initialized with a list of different words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure. Implement the MagicDictionary class: MagicDictionary() Initializes the object. void buildDict(String[] dictionary) Sets the data structure with an array of distinct strings dictionary. bool search(String searchWord) Returns true if you can change exactly one character in searchWord to match any string in the data structure, otherwise returns false. Example 1: Input ["MagicDictionary", "buildDict", "search", "search", "search", "search"] [[], [["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]] Output [null, null, false, true, false, false] Explanation MagicDictionary magicDictionary = new MagicDictionary(); magicDictionary.buildDict(["hello", "leetcode"]); magicDictionary.search("hello"); // return False magicDictionary.search("hhllo"); // We can change the second 'h' to 'e' to match "hello" so we return True magicDictionary.search("hell"); // return False magicDictionary.search("leetcoded"); // return False Constraints: 1 <= dictionary.length <= 100 1 <= dictionary[i].length <= 100 dictionary[i] consists of only lower-case English letters. All the strings in dictionary are distinct. 1 <= searchWord.length <= 100 searchWord consists of only lower-case English letters. buildDict will be called only once before search. At most 100 calls will be made to search.

Explanation

Here's a breakdown of the approach, complexities, and the Python code for the MagicDictionary problem:

• High-Level Approach:

  • Store the dictionary words in a list.
  • For each search word, iterate through the dictionary and check if exactly one character difference exists.
  • A helper function efficiently counts the differences between two strings.

• Complexity:

  • Runtime: buildDict is O(N), where N is the number of words in the dictionary. search is O(N * L), where N is the number of words in the dictionary and L is the length of the search word (and the words in the dictionary which have matching lengths).
  • Storage: O(N), where N is the number of words in the dictionary (to store the dictionary words).

Code

    class MagicDictionary:

    def __init__(self):
        self.dictionary = []

    def buildDict(self, dictionary: list[str]) -> None:
        self.dictionary = dictionary

    def search(self, searchWord: str) -> bool:
        for word in self.dictionary:
            if len(word) == len(searchWord):
                diff_count = 0
                for i in range(len(word)):
                    if word[i] != searchWord[i]:
                        diff_count += 1
                if diff_count == 1:
                    return True
        return False