526. Beautiful Arrangement

Problem Link : https://leetcode.com/problems/beautiful-arrangement/
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Complexity : Time-O(2^N), Memory-O(N), TimeTaken-15min
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Logic :
I used an array to simplify the complexity.
Suppose array as the num from 1-n whenever we pick a num from array we put true there
and send it next traversal and start to choose next num from start of the array and if 
that index is already taken we try next index and in that way we create/ generate next perm
and increase the counter
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class Solution {
    int count=0;
    public int countArrangement(int n) {
        traverse(new boolean[n], 0, 0);
        return count;
    }
    private void traverse(boolean[] taken, int i, int len){
        if(len==taken.length){
            ++count;
            return;
        }
        for(int i1=0;i1<taken.length;i1++){
            if(!taken[i1] && isBeautiful(i+1, i1+1)){
                taken[i1]=true;
                traverse(taken, i+1, len+1);
                taken[i1]=false;
            }
        }
    }
    private boolean isBeautiful(int i, int j){
        return i%j==0 || j%i==0;
    }
}