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 = 2
Output: 2
Explanation:
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 = 1 Output: 1
Constraints:
1 <= n <= 15struct Solution;
impl Solution {
fn count_arrangement(n: i32) -> i32 {
let mut res = 0;
let mut nums = (1..=n).collect();
let n = n as usize;
Self::dfs(0, &mut nums, &mut res, n);
res
}
fn dfs(start: usize, nums: &mut Vec<i32>, all: &mut i32, n: usize) {
if start == n {
*all += 1;
} else {
for i in start..n {
if nums[i] % (start as i32 + 1) == 0 || (start as i32 + 1) % nums[i] == 0 {
nums.swap(start, i);
Self::dfs(start + 1, nums, all, n);
nums.swap(start, i);
}
}
}
}
}
#[test]
fn test() {
let n = 2;
let res = 2;
assert_eq!(Solution::count_arrangement(n), res);
}