<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> circularPermutation(int n, int start) {

        // we need to solve it using grey code (consecutive numbers diff by 1 bit)
        // To convert number to its grey code. 
        // first digit should be same, second digit in grey code is xor of 1st and 2nd digits, ith digit is xor of ith, i-1th digit in    binary representation
        // i>>1 is i/(2power1)
        // i ^ (i/2) => i ^ (i>>1) gives grey code representation
        // since it should be started with start, we xor with start.


        int i, grey, val;
        vector<int>res;
        for(i=0; i< 1<<n; i++) {
            grey = i ^ i>>1;
            val = grey ^ start;
            res.push_back(val);
        }

        return res;
    }
};
19000