feat: [Week 06-3] solve design-add-and-search-words-data-structure

Author: Chaedie

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

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+"""
+풀이를 보고 공부했습니다.
+
+Solution:
+    set을 이용,
+    1) addWord 에선 add
+    2) search 에선 .을 포함한 모든 글자가 동일한 단어가 있는지 확인
+
+search 
+    Time: O(n * w)
+    Space: O(1)
+"""
+
+
+class WordDictionary:
+
+    def __init__(self):
+        self.root = set()
+
+    def addWord(self, word: str) -> None:
+        self.root.add(word)
+
+    def search(self, word: str) -> bool:
+        for candidate in self.root:
+            if len(candidate) != len(word):
+                continue
+            if all(w == c or w == "." for w, c in zip(word, candidate)):
+                return True
+        return False
+
+
+"""
+풀이를 보고 공부했습니다.
+
+Solution: Trie - 달레의 코드
+"""
+
+
+class WordDictionary:
+
+    def __init__(self):
+        self.root = {"$": True}
+
+    def addWord(self, word: str) -> None:
+        node = self.root
+        for ch in word:
+            if ch not in node:
+                node[ch] = {"$": False}
+            node = node[ch]
+        node["$"] = True
+
+    def search(self, word: str) -> bool:
+        def dfs(node, idx):
+            if idx == len(word):
+                return node["$"]
+
+            ch = word[idx]
+            if ch in node:
+                return dfs(node[ch], idx + 1)
+            if ch == ".":
+                if any(dfs(node[k], idx + 1) for k in node if k != "$"):
+                    return True
+            return False
+
+        return dfs(self.root, 0)
+
+
+"""
+풀이를 보고 공부했습니다.
+
+Solution: Trie with TrieNode - NeetCode
+"""
+
+
+class TrieNode:
+    def __init__(self):
+        self.children = {}  # a: TrieNode
+        self.word = 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.word = True
+
+    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.word
+
+        return dfs(0, self.root)