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

Author: dusunax

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

@@ -0,0 +1,67 @@
+'''
+# 211. Design Add and Search Words Data Structure
+
+use trie to perform add and search operations on words.
+use recursive dfs to search for words with "." wildcards.
+
+## Time and Space Complexity
+
+### addWord
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through each character of the word. = O(n)
+
+### SC is O(n):
+- storing the word in the trie. = O(n)
+
+### search
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- dfs iterates through each character of the word. = O(n)
+- if char is "."(wildcard) dfs iterates through all children of the current node. = O(n)
+
+> recursion for the wildcard could involve exploring several paths.
+> but time complexity is bounded by the length of the word, so it's still O(n).
+
+### SC is O(n):
+- length of the word as recursive call stack. = O(n)
+'''
+class WordDictionary:
+    def __init__(self):
+        self.trie = {}
+
+    def addWord(self, word: str) -> None:
+        node = self.trie
+        for char in word: # TC: O(n)
+            if char not in node:
+                node[char] = {}
+            node = node[char] 
+        node["$"] = True
+
+    def search(self, word: str) -> bool:
+        def dfs(node, i) -> bool:
+            if not isinstance(node, dict):
+                return False
+            if i == len(word):
+                return "$" in node if isinstance(node, dict) else False 
+            char = word[i]
+
+            if char == ".":
+                for child in node.values():
+                    if dfs(child, i + 1):
+                        return True
+                return False
+            else:
+                return dfs(node[char], i+1) if char in node else False
+        
+        return dfs(self.trie, 0)