526. Beautiful Arrangement (Medium)

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 $$i_{th}$$ position (1 ≤ i ≤ N) in this array:

1. The number at the $$i_{th}$$ position is divisible by i.
2. i is divisible by the number at the $$i_{th}$$ 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:

1. N is a positive integer and will not exceed 15.

Solution: Backtracking

Guess the reason is:

1. Init, every number are in the right position: nums[i] == i; (index start from 1....)
2. Then, if nums[i]%n==0 || n%nums[i]==0 is true, then i % n || nums[n] % i is also true, both situation indicates safe to swap.
3. And the swap keeps the property of that, so its no need to check the "(vs[n-1]%(i+1))==0 || (i+1)%vs[n-1])". Just guess, hope someone can give a entire proof.

I think if you change for loop condition to for (int i = n-1; i>=0; i--) it makes me easier to understand cause it's actually verifying if the number could be divided by index while doing the permutation

version 1: 6ms

class Solution {
public:
    int countArrangement(int N) {
        vector<int> v;
        for (int i = 1; i <= N; ++i) v.push_back(i);
        return counts(N, v);
    }
    int counts(int n, vector<int>& v) {
        if (n <= 0) return 1;
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (v[i]%n==0 || n%v[i]==0) {
                swap(v[i], v[n-1]);
                cnt += counts(n-1, v);
                swap(v[i], v[n-1]);
            }
        }
        return cnt;
    }
};

version 2: 166ms

class Solution {
    void helper(int N, int num, vector<bool>& used) {
        if (num > N) { ++cnt; return; }
        for (int i = 1; i <= N; ++i) {
            if (!used[i] && (i%num == 0 || num%i == 0)) {
                used[i] = true;
                helper(N, num+1, used);
                used[i] = false;
            }
        }

    }
public:
    int countArrangement(int N) {
        if (N == 0) return 0;
        vector<bool> used(N+1, false);
        helper(N, 1, used);
        return cnt;
    }
private:
    int cnt = 0;
};