from lib_misc import permutations


def euler_024_library():
    from itertools import permutations
    return int("".join(list(permutations("0123456789"))[1000000 - 1]))


def nth(generator, n):
    for i in range(n):
        next(generator)
    return next(generator)


def nth_permutation(iterable, n):
    """ Returns the nth permutation of the iterable. """

    def iterate_permutations(accumulator, iterable, count=[0]):
        if not iterable:
            if count[0] == n:
                return accumulator
            count[0] += 1
            return False
        for i in range(len(iterable)):
            elem = iterable[i:i + 1]
            rest = iterable[:i] + iterable[i + 1:]
            p = iterate_permutations(accumulator + elem, rest)
            if p:
                return p
        return False

    return iterate_permutations([], iterable)


def euler_024():
    """ My first solution used a generator to get all permutations till
    the desired one. My improved solution uses a function that returns only
    the nth iteration which is the same algorithm, but does not required to
    return all the permutations that are discarded anyway. """
    return int("".join(nth_permutation(list("0123456789"), 1000000 - 1)))
    #  return int(nth(permutations("0123456789"), 1000000 - 1))


if __name__ == "__main__":
    print("e024.py: {}".format(euler_024()))
    assert(euler_024() == 2783915460)