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 by i.
• i is divisible by perm[i].

Given an integer n, return the number of the beautiful arrangements that you can construct.

• new solution update 2022.03.07

class Solution {
    int dp[1<<16];
    int backTracking(int& n, int perm, int mask) {
        if(perm == n + 1) {
            return 1;
        }
        if(dp[mask]!=-1) return dp[mask];
        dp[mask] = 0;
        for(int i = 1; i <= n; i++) {
            if(mask & (1<<i)) continue;
            if(perm % i == 0 or i % perm == 0) {
                dp[mask] += backTracking(n, perm + 1, mask | (1<<i));
            }
        }
        return dp[mask];
    }
public:
    int countArrangement(int n) {
        memset(dp,-1,sizeof(dp));
        return backTracking(n, 1, 0);
    }
};

class Solution {
    vector<int> perm;
    int sol(int n) {
        if(!n) return 1;
        int res(0);
        for(int i = 0; i < n; i++) {
            if(!(perm[i] % n) || !((n % perm[i]))) {
                swap(perm[i], perm[n - 1]);
                res += sol(n - 1);
                swap(perm[i], perm[n - 1]);
            }
        }
        return res;
    }
public:
    int countArrangement(int n) {
        for(int i = 0; i < n; i++) perm.push_back(i + 1);
        return n < 4 ? n : sol(n);
    }
};