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
function circularPermutation(n: number, start: number): number[] {
    const ans: number[] = [];
    for (let i = 0; i < 1 << n; ++i) {
        ans.push(i ^ (i >> 1) ^ start);
    }
    return ans;
}
