Algorithm
Problem Name: 526. 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 byi.iis divisible byperm[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 <= 15
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
var countArrangement = function(N) {
let used = Array(N + 1).fill(false)
let res = 0
function backtrack(curIdx) {
if (curIdx === 0) return res++
for (let i = 1; i < = N; i++) {
if (used[i]) continue
if (i % curIdx === 0 || curIdx % i === 0) {
used[i] = true
backtrack(curIdx - 1)
used[i] = false
}
}
}
backtrack(N)
return res
};
Input
n = 2
Output
2
#2 Code Example with Python Programming
Code -
Python Programming
memo = {}
class Solution:
def countArrangement(self, N, arr = None):
if not arr: arr = tuple(range(1, N + 1))
if (N, arr) in memo or N == 1: return N == 1 and 1 or memo[(N, arr)]
memo[(N, arr)] = sum([self.countArrangement(N-1, arr[:j]+arr[j + 1:]) for j in range(len(arr)) if arr[j]%N==0 or N%arr[j]==0])
return memo[(N, arr)]
Input
n = 2
Output
2