LeetCode - Design Add and Search Words Data Structure

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

• WordDictionary() Initializes the object.
• void addWord(word) Adds word to the data structure, it can be matched later.
• bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.

 

Example:

Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]

Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True

 

Constraints:

• 1 <= word.length <= 25
• word in addWord consists of lowercase English letters.
• word in search consist of '.' or lowercase English letters.
• There will be at most 2 dots in word for search queries.
• At most 104 calls will be made to addWord and search.

 

My Solution:

class WordDictionary:

    def __init__(self):
        self.words = set()


    def addWord(self, word: str) -> None:
        self.words.add(word)


    def search(self, word: str) -> bool:
        idx_1 = word.find('.')
        if idx_1 == -1:  # word contains no '.'
            return word in self.words
        else:  # word contains at least one '.'
            idx_2 = word[idx_1 + 1:].find('.')
            idx_2 = idx_1 + 1 + idx_2

            for w in self.words:
                if len(w) != len(word):
                    continue

                if word[:idx_1] == w[:idx_1] and word[idx_1 + 1:idx_2] == w[idx_1 + 1:idx_2] and word[idx_2 + 1:] == w[idx_2 + 1:]:
                    return True
        
            return False


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)

 

 

ChatGPT's Solution:

class TrieNode:
    def __init__(self):
        # Each node stores a dictionary of children and a boolean for end of word.
        self.children = {}
        self.is_end_of_word = False


class WordDictionary:
    def __init__(self):
        self.root = TrieNode()

    def addWord(self, word: str) -> None:
        node = self.root
        for char in word:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
        node.is_end_of_word = True

    def search(self, word: str) -> bool:
        return self._search_in_node(word, 0, self.root)
    
    def _search_in_node(self, word: str, index: int, node: TrieNode) -> bool:
        if index == len(word):
            return node.is_end_of_word
        char = word[index]
        if char == '.':
            for child in node.children.values():
                if self._search_in_node(word, index + 1, child):
                    return True
        elif char in node.children:
            return self._search_in_node(word, index + 1, node.children[char])
        return False


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)