Bruteforce word with a specified separator

find_words.py

    def find_words(instring, words, prefix=''):
        """ Quick and dirty snippet to split string without space
            to a list of known words.
            Used to pass over a bug of Snap!"""
        if not instring:
            return []

        if (not prefix) and (instring in words):
            return [instring]
        prefix, suffix = prefix + instring[0], instring[1:]
        solutions = []
        # Case 1: prefix in solution
        if prefix in words:
            try:
                solutions.append([prefix] + __find_words__(suffix, words, ''))
            except ValueError:
                pass
        # Case 2: prefix not in solution
        try:
            solutions.append(__find_words__(suffix, words, prefix))
        except ValueError:
            pass
        if solutions:
            return sorted(solutions,
                          key=lambda solution: [len(word) for word in solution],
                          reverse=True)[0]
        else:
            return instring