Is Implement Magic Dictionary easy or hard?

Implement Magic Dictionary is rated Medium because the brute-force idea is straightforward but the optimal solution requires recognizing the wildcard pattern trick. Understanding how to convert a one-character difference rule into a hash lookup is the key insight.

Implement Magic Dictionary Python/Java solution

Python and Java implementations typically build a HashMap or dictionary where keys are wildcard patterns like 'h*llo'. During search, the algorithm generates patterns for the query word and checks the map for matches while ensuring exactly one character difference.

How to solve Implement Magic Dictionary in O(n)?

The practical optimal solution is O(n*m) preprocessing and O(m) search using a hash map of wildcard patterns. Each word produces m patterns by replacing one character with '*'. During search, matching patterns quickly identify words that differ by exactly one character.

What is the best approach for Implement Magic Dictionary?

The hash map wildcard pattern approach is the most efficient and commonly expected solution. Each dictionary word generates patterns by replacing one character with '*', which are stored in a hash table. During search, the query word generates the same patterns and checks for matches in O(m) time after O(n*m) preprocessing.

Is Implement Magic Dictionary asked at Google/Amazon/Meta?

Design-style dictionary problems and string mutation checks appear frequently in interviews at companies like Google, Amazon, and Meta. Variants often test hash table usage, string manipulation, and data structure design concepts similar to Magic Dictionary.

What data structure is used in Implement Magic Dictionary?

The most common solution uses a hash table that maps wildcard string patterns to counts or lists of dictionary words. Alternatives may use tries or direct string comparison, but the hash map pattern technique provides the best balance of simplicity and performance.

What is the time complexity of Implement Magic Dictionary?

Using the optimized hash map pattern approach, building the dictionary takes O(n*m) time where n is the number of words and m is the word length. Each search operation runs in O(m) because only m wildcard patterns are generated and checked in a hash table.

Description Solution Video 14:27

676. Implement Magic Dictionary

Medium58.4% acceptance2 companies

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: