# Suppose you have N integers from 1 to N. We define a beautiful arrangement as an array that is constructed by these N numbers successfully if one of the following is true for the ith position (1 <= i <= N) in this array:
#
# The number at the ith position is divisible by i.
# i is divisible by the number at the ith position.
# Now given N, how many beautiful arrangements can you construct?
#
# Example 1:
# Input: 2
# Output: 2
# Explanation:
#
# The first beautiful arrangement is [1, 2]:
#
# Number at the 1st position (i=1) is 1, and 1 is divisible by i (i=1).
#
# Number at the 2nd position (i=2) is 2, and 2 is divisible by i (i=2).
#
# The second beautiful arrangement is [2, 1]:
#
# Number at the 1st position (i=1) is 2, and 2 is divisible by i (i=1).
#
# Number at the 2nd position (i=2) is 1, and i (i=2) is divisible by 1.
# Note:
# N is a positive integer and will not exceed 15.


# My X is the set of still available numbers. Gets accepted in about 230 ms:
def countArrangement(self, N):
    def count(i, X):
        if i == 1:
            return 1
        return sum(count(i - 1, X - {x})
                   for x in X
                   if x % i == 0 or i % x == 0)
    return count(N, set(range(1, N + 1)))
# Note that my i goes downwards, from N to 1. Because position i = 1 can hold any
# number, so I don't even have to check whether the last remaining number fits
# there. Also, position i = 2 happily holds every second number and i = 3 happily
# holds every third number, so filling the lowest positions last has a relatively
# high chance of success. In other words, it's relatively hard to end up with dead ends this way.

# If I go upwards (from 1 to N), it takes about 2300 ms:

def countArrangement(self, N):
    def count(i, X):
        if i > N:
            return 1
        return sum(count(i + 1, X - {x})
                   for x in X
                   if x % i == 0 or i % x == 0)
    return count(1, set(range(1, N + 1)))
