Circular Permutation in Binary Representation
<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>
Bit Manipulation; Math; Backtracking
class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        int[] g = new int[1 << n];
        int j = 0;
        for (int i = 0; i < 1 << n; ++i) {
            g[i] = i ^ (i >> 1);
            if (g[i] == start) {
                j = i;
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = j; i < j + (1 << n); ++i) {
            ans.add(g[i % (1 << n)]);
        }
        return ans;
    }
}