Python Short Brute Force 85% Time
```\nclass Solution:\n    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:\n        # Check if current word can be made from smaller wo
```\nclass Solution:\n    def findAllConcatenatedWordsInADict(self, words: List[str]) -> List[str]:\n        # Check if current word can be made from smaller words\n        def check(word):\n            if word in s:\n                return True\n            # If prefix is seen, recurse on rest of word\n            return any(check(word[i:]) for i in range(1, len(word)) if word[:i] in s)\n        \n        s = set()\n        res = []\n        for word in sorted(words, key=lambda x:len(x)):\n            if check(word):\n                res.append(word)\n            s.add(word)\n        return res
['Recursion', 'Python', 'Python3']