LC 211 - Design Add and Search Words Data Structure

Question

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.

Links

Question here and solution here

Solution

concept

The only difficult part is to use DFS search for the wild card .

code

class TrieNode():
    def __init__(self):
        self.children = {} # a : TrieNode
        self.end_of_word = False

class WordDictionary:

    def __init__(self):
        self.root = TrieNode()
        
    def addWord(self, word: str) -> None:
        cur = self.root
        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()
            cur = cur.children[c]
        cur.end_of_word = True

    def search(self, word: str) -> bool:
        def dfs(j, root):
            cur = root

            for i in range(j, len(word)):
                c = word[i]
                if c == ".":
                    for child in cur.children.values():
                        if dfs(i + 1, child):
                            return True
                    return False
                else:
                    if c not in cur.children:
                        return False
                    cur = cur.children[c]
            return cur.end_of_word
        
        return dfs(0, self.root)

Complexity

time: $O(n)$
space: $O(t + n)$