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211. Design Add and Search Words Data Structure

Difficulty Topics

Description

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 3 dots in word for search queries.
  • At most 104 calls will be made to addWord and search.

Solution

design-add-and-search-words-data-structure.py
class TrieNode:
    def __init__(self):
        self.children = [None] * 26
        self.end = False

class WordDictionary:

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

    def addWord(self, word: str) -> None:
        curr = self.root
        n = len(word)
        for i in range(n):
            index = ord(word[i]) - ord('a')
            if not curr.children[index]:
                curr.children[index] = TrieNode()

            curr = curr.children[index]

        curr.end = True

    def search(self, word: str) -> bool:
        self.res = False

        self.dfs(self.root, word)

        return self.res

    def dfs(self, node, word):
        if not word:
            if node.end: self.res = True

            return

        if self.isLastNode(node): return

        if word[0] == ".":
            for i in range(26):
                if node.children[i]:
                    self.dfs(node.children[i], word[1:])

        else:
            index = ord(word[0]) - ord('a')
            if node.children[index]:
                self.dfs(node.children[index], word[1:])


    def isLastNode(self, node):
        for i in range(26):
            if node.children[i]: return False

        return True








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