<p>Given 2 integers <code>n</code> and <code>start</code>. Your task is return <strong>any</strong> permutation <code>p</code>&nbsp;of <code>(0,1,2.....,2^n -1) </code>such that :</p>

<ul>
	<li><code>p[0] = start</code></li>
	<li><code>p[i]</code> and <code>p[i+1]</code>&nbsp;differ by only one bit in their binary representation.</li>
	<li><code>p[0]</code> and <code>p[2^n -1]</code>&nbsp;must also differ by only one bit in their binary representation.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2, start = 3
<strong>Output:</strong> [3,2,0,1]
<strong>Explanation:</strong> The binary representation of the permutation is (11,10,00,01). 
All the adjacent element differ by one bit. Another valid permutation is [3,1,0,2]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, start = 2
<strong>Output:</strong> [2,6,7,5,4,0,1,3]
<strong>Explanation:</strong> The binary representation of the permutation is (010,110,111,101,100,000,001,011).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 16</code></li>
	<li><code>0 &lt;= start&nbsp;&lt;&nbsp;2 ^ n</code></li>
</ul>
class Solution {
private:
bool isBitDifference1(int A, int B, int n) {
    int count = 0;
    for (int i = 0; i <=n; i++) {
        if ((A & (1 << i)) != (B & (1 << i))) {
            count++;
        }
    }
   return count==1;
}
public:
    vector<int> circularPermutation(int n, int start) {
        vector<int> ans(1<<n,-1);
        unordered_set<int> used;

        ans[0] = start;

        int l=1, r=(1<<n)-1;
        bool c=1;
        used.insert(start);
        
        while(l<=r) {
            int lastVal=-1;
            int idx=-1;
            if(!c){
                lastVal = ans[l-1];
                idx=l;
                l++;
            }else {
                lastVal = (r+1)==(1<<n) ? ans[0] : ans[r+1];
                idx=r;
                r--;
            }
            int xx = 0;
            while(xx <= ((1<<n)-1)) {
                if(!isBitDifference1(lastVal,lastVal ^ xx, n) || used.count(lastVal^xx)){
                    xx++;
                } else{
                    break;
                }
            }
            used.insert(lastVal^xx);
            ans[idx] = lastVal^xx;
            c = !c;
        }
        return ans;
    }
};
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