"""
211. Design Add and Search Words Data Structure
Difficulty: Medium
https://leetcode.com/problems/design-add-and-search-words-data-structure/

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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 10^4 calls will be made to addWord and search.
"""


class TrieNode:
    def __init__(self):
        self.children = {}
        self.endOfWord = 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.endOfWord = True

    # Leverage Recursion
    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.endOfWord

        return dfs(0, self.root)


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