## Description

https://leetcode.com/problems/beautiful-arrangement/

Suppose you have `n`

integers labeled `1`

through `n`

. A permutation of those `n`

integers `perm`

(**1-indexed**) is considered a **beautiful arrangement** if for every `i`

(`1 <= i <= n`

), **either** of the following is true:

`perm[i]`

is divisible by`i`

.`i`

is divisible by`perm[i]`

.

Given an integer `n`

, return *the number of the beautiful arrangements that you can construct*.

**Example 1:**

Input:n = 2Output:2Explanation:The first beautiful arrangement is [1,2]: - perm[1] = 1 is divisible by i = 1 - perm[2] = 2 is divisible by i = 2 The second beautiful arrangement is [2,1]: - perm[1] = 2 is divisible by i = 1 - i = 2 is divisible by perm[2] = 1

**Example 2:**

Input:n = 1Output:1

**Constraints:**

`1 <= n <= 15`

## Explanation

Use the backtracking approach to find all the combinations where each position meets nums[i] is divisible by i or i is divisible by nums[i] condition.

## Python Solution

```
class Solution:
def countArrangement(self, n: int) -> int:
numbers = [i for i in range(1, n + 1)]
results = []
visited = set()
self.helper(results, numbers, [], visited)
return len(results)
def helper(self, results, numbers, combination, visited):
if len(combination) == len(numbers):
results.append(list(combination))
return
for num in numbers:
if num in visited:
continue
if num % (len(combination) + 1) != 0 and (len(combination) + 1) % num != 0:
continue
combination.append(num)
visited.add(num)
self.helper(results, numbers, combination, visited)
combination.pop()
visited.remove(num)
```

- Time Complexity: O(N).
- Space Complexity: O(N).