Skip to main content

Design Add and Search Words Data Structure

Problem Statement​

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.

Leetcode Link

Code​

Python
class WordDictionary:

    def __init__(self):

        # Initialize your data structure here.
        self.children = [None]*26
        self.isEndOfWord = False


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

        # Adds a word into the data structure.
        curr = self
        for c in word:
            if curr.children[ord(c) - ord('a')] == None:
                curr.children[ord(c) - ord('a')] = WordDictionary()
            curr = curr.children[ord(c) - ord('a')]

        curr.isEndOfWord = True;


    def search(self, word: str) -> bool:

        # Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
        curr = self
        for i in range(len(word)):
            c = word[i]
            if c == '.':
                for ch in curr.children:
                    if ch != None and ch.search(word[i+1:]): return True
                return False

            if curr.children[ord(c) - ord('a')] == None: return False
            curr = curr.children[ord(c) - ord('a')]

        return curr != None and curr.isEndOfWord