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

SOLUTION:

class Node:
    def __init__(self, val=None, children=[], end=False):
        self.val = val
        self.children = children
        self.end = end

class WordDictionary:

    def __init__(self):
        self.root = Node(val=None, children=[])

    def addWord(self, word: str) -> None:
        n = len(word)
        curr = self.root
        for i, c in enumerate(word):
            found = False
            for node in curr.children:
                if node.val == c:
                    curr = node
                    found = True
                    break
            if not found:
                newcurr = Node(val=c, children=[])
                curr.children.append(newcurr)
                curr = newcurr
        curr.end = True

    def search(self, word: str) -> bool:
        n = len(word)
        paths = [(self.root, 0)]
        while len(paths) > 0:
            curr, i = paths.pop()
            if i == n and curr.end:
                return True
            if curr.children and i < n:
                for c in curr.children:
                    if word[i] == "." or c.val == word[i]:
                        paths.append((c, i + 1))
        return False

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