<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 {
public:
    vector<int>find(int n){
        if(n==1){
            return {0,1};
        }
        vector<int>res=find(n-1);
        int s=res.size();
        vector<int>ans;
        for(int i=0;i<s;i++){
            ans.push_back(res[i]);
        }
        for(int i=s-1;i>=0;i--){
            int x=(res[i]/2)+1;
            ans.push_back(res[i]+pow(2,n-1));
        }
        return ans;
    }
    vector<int> circularPermutation(int n, int start) {
        vector<int>ans=find(n);
        int s=ans.size();
        int i=0;
        while(i<s){
            if(ans[i]==start)break;
            i++;
        }
        if(i==0)return ans;
        reverse(ans.begin(),ans.begin()+i);
        reverse(ans.begin()+i,ans.end());
        reverse(ans.begin(),ans.end());
        return ans;
    }
};
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