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 byi
.i
is divisible byperm[i]
.
Given an integer n
, return the number of the beautiful arrangements that you can construct.
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
Input: n = 1 Output: 1
1 <= n <= 15
impl Solution {
pub fn count_arrangement(n: i32) -> i32 {
Self::backtracking(n, 1, 0)
}
fn backtracking(n: i32, i: i32, bitmask: i32) -> i32 {
if i > n {
return 1;
}
(1..=n)
.filter(|p| (bitmask >> p) & 1 == 0 && (p % i == 0 || i % p == 0))
.map(|p| Self::backtracking(n, i + 1, bitmask | (1 << p)))
.sum()
}
}