[LeetCode] 526. Beautiful Arrangement

Problem

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 ith position (1 <= i <= N) in this array:

The number at the ith position is divisible by i.
i is divisible by the number at the ith 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:
N is a positive integer and will not exceed 15.

Solution

class Solution {
    int count = 0;
    public int countArrangement(int N) {
        //for each position, fill with 1-N using dfs
        //when position moved to N+1, we found another arrangement
        boolean[] used = new boolean[N+1];
        dfs(1, N, used);
        return count;
    }
    private void dfs(int index, int len, boolean[] used) {
        if (index > len) {
            count++;
            return;
        }
        for (int i = 1; i <= len; i++) {
            if (!used[i] && (i%index == 0 || index%i == 0)) {
                used[i] = true;
                dfs(index+1, len, used);
                used[i] = false;
            }
        }
    }
}