Word Break Solution

Author: jinbeom

word-break/kayden.py

@@ -0,0 +1,25 @@
+# 시간복잡도: ?
+# 공간복잡도: O(S)
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        memo = {}
+        wordSet = set(wordDict)
+
+        def dfs(idx):
+            if idx in memo:
+                return memo[idx]
+
+            if idx == len(s):
+                return True
+
+            for word in wordSet:
+                l = len(word)
+                if s[idx:idx + l] == word:
+                    if dfs(idx + l):
+                        memo[idx] = True
+                        return True
+
+            memo[idx] = False
+            return False
+
+        return dfs(0)