How do you solve Implement Magic Dictionary on LeetCode?

Implement Magic Dictionary (LeetCode #676) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Using a Trie can significantly reduce the number of comparisons needed.

What is the brute force solution for Implement Magic Dictionary?

The brute force approach for Implement Magic Dictionary is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each word in the dictionary against the search word by changing each character one at a time. If we find a match, we return true; otherwise, we return false after checking all words. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Implement Magic Dictionary?

Implement Magic Dictionary uses the following concepts: Hash Table, String, Depth-First Search, Design, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Implement Magic Dictionary?

Explain your thought process clearly while coding. Consider edge cases and how they affect your solution. Practice building and using data structures like Trie before the interview.

What are common mistakes when solving Implement Magic Dictionary?

Not considering length mismatches early in the brute-force approach. Failing to handle edge cases where no words are present.

#676

Implement Magic Dictionary

Medium

Hash TableStringDepth-First SearchDesignTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

💡

Intuition

Time O(n * m)Space O(n * m)

Using a Trie allows us to efficiently store the words and check for matches with a single character change. This structure helps in reducing the number of comparisons needed by leveraging the prefix nature of words.

⚙️

Algorithm

3 steps

1Step 1: Build a Trie from the list of words in the dictionary.
2Step 2: For each character in the searchWord, traverse the Trie while allowing one character to differ.
3Step 3: If we reach the end of a valid path in the Trie with exactly one character changed, return true; otherwise, return false.

solution.py33 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.is_end = False
5
6class MagicDictionary:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def buildDict(self, dictionary):
11        for word in dictionary:
12            node = self.root
13            for char in word:
14                if char not in node.children:
15                    node.children[char] = TrieNode()
16                node = node.children[char]
17            node.is_end = True
18
19    def search(self, searchWord):
20        return self._search(self.root, searchWord, 0, False)
21
22    def _search(self, node, word, index, modified):
23        if index == len(word):
24            return modified and node.is_end
25        char = word[index]
26        for key in node.children:
27            if key == char:
28                if self._search(node.children[key], word, index + 1, modified):
29                    return True
30            elif not modified:
31                if self._search(node.children[key], word, index + 1, True):
32                    return True
33        return False

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words and m is the maximum length of the words. This is because we traverse each word in the dictionary to build the Trie and then perform a search that can potentially check all characters in the Trie. The space complexity is O(n * m) due to the storage of the Trie structure.

1Using a Trie can significantly reduce the number of comparisons needed.
2Understanding character differences is crucial for this problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.