Design add and search words data structure

class TrieNode:
    def __init__(self):
        self.children = [None] * 26
        self.isEnd = False

    def insert(self, word: str) -> None:
        node = self
        for ch in word:
            ch = ord(ch) - ord('a')
            if not node.children[ch]:
                node.children[ch] = TrieNode()
            node = node.children[ch]
        node.isEnd = True


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

    def addWord(self, word: str) -> None:
        self.trieRoot.insert(word)

    def search(self, word: str) -> bool:
        def dfs(index: int, node: TrieNode) -> bool:
            if index == len(word):
                return node.isEnd
            ch = word[index]
            if ch != '.':
                child = node.children[ord(ch) - ord('a')]
                if child is not None and dfs(index + 1, child):
                    return True
            else:
                for child in node.children:
                    if child is not None and dfs(index + 1, child):
                        return True
            return False

        return dfs(0, self.trieRoot)

Design Add and Search Words Data Structure

Difficulty: Medium

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.