Beautiful arrangement

Time: O(N!); Space: O(N); medium

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: 1. The number at the ith position is divisible by i. 2. 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.

Example 2:

Input: 3

Output: 3

Note:

  • N is a positive integer and will not exceed 15.

[4]:

class Solution1(object):
    """
    Time: O(N!)
    Space: O(N)
    """
    def countArrangement(self, N):
        """
        :type N: int
        :rtype: int
        """
        def countArrangementHelper(n, arr):
            if n <= 0:
                return 1
            count = 0
            for i in range(n):
                if arr[i] % n == 0 or n % arr[i] == 0:
                    arr[i], arr[n-1] = arr[n-1], arr[i]
                    count += countArrangementHelper(n - 1, arr)
                    arr[i], arr[n-1] = arr[n-1], arr[i]
            return count

        return countArrangementHelper(N, list(range(1, N+1)))

[5]:

s = Solution1()
N = 2
assert s.countArrangement(N) == 2
N = 3
assert s.countArrangement(N) == 3