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.
class Solution:
def countArrangement(self, N: int) -> int:
self.ans = 0
self.mark = [False for i in range(N+1)]
self.dfs(N, self.mark, [])
return self.ans
def dfs(self, N, mark, path):
if len(path) == N:
self.ans += 1
for n in range(1, N+1):
index = len(path) + 1
if mark[n] or (n % index != 0 and index % n != 0):
continue
mark[n] = True
self.dfs(N, self.mark, path + [n])
mark[n] = False