design-add-and-search-words-data-structure solution

Author: jungsiroo

design-add-and-search-words-data-structure/jungsiroo.py

@@ -0,0 +1,60 @@
+class Node:
+    def __init__(self, char=None):
+        self.is_end = False
+        self.children = {}
+
+class WordDictionary:
+    """
+    트라이를 활용
+    기본 문제와 다른 점은 search 할 때 "." 이 포함되어있는 것
+
+    따라서 . 이라면 해당 노드의 모든 자식을 모두 조회해야됨
+
+    addWord
+    w = len(word)
+    Tc : O(w)
+    Sc : O(w)
+
+    search
+    최악의 경우 case : xa, xb, xc, xd, .... xz 로 x의 다음 글자가 a~z일 때
+    search("x.") 라면 26개 모두를 찾아봐야됨
+    Tc : O(26^w)
+    Sc : O(w) (글자 길이만큼 재귀를 호출하기에)
+
+    """
+    def __init__(self):
+        self.root = Node()
+
+    def addWord(self, word: str) -> None:
+        now = self.root
+
+        for ch in word:
+            if ch not in now.children:
+                now.children[ch] = Node(ch)
+            now = now.children[ch] 
+        now.is_end = True
+
+    def search(self, word: str) -> bool:
+        def _recur_search(node, index):
+            if index == len(word):
+                return node.is_end
+            
+            if word[index] == ".":
+                for ch in node.children:
+                    if _recur_search(node.children[ch], index+1):
+                        return True
+            
+            if word[index] in node.children:
+                return _recur_search(node.children[word[index]], index+1)
+            return False 
+        
+        return _recur_search(self.root, 0)
+            
+        
+
+
+# Your WordDictionary object will be instantiated and called as such:
+# obj = WordDictionary()
+# obj.addWord(word)
+# param_2 = obj.search(word)
+