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+"""
+211. Design Add and Search Words Data Structure
+Difficulty: Medium
+https://leetcode.com/problems/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 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)