<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:
    def circularPermutation(self, n: int, start: int) -> List[int]:
        def grayCode(n):
            if n == 0:
                return [0]
            result = []
            gray_code_sequence = grayCode(n - 1)
            result.extend(gray_code_sequence)
            for i in range(len(gray_code_sequence) - 1, -1, -1):
                result.append((1 << (n - 1)) | gray_code_sequence[i])
            return result
        
        def rotate_start(sequence, start):
            start_index = sequence.index(start)
            return sequence[start_index:] + sequence[:start_index]
        
        gray_codes = grayCode(n)
        
        # Rotate the sequence to start from the given start value
        rotated_codes = rotate_start(gray_codes, start)
        
        # Convert the Gray codes to decimal and return
        return rotated_codes

# Example usage:
solution = Solution()
print(solution.circularPermutation(2, 3))  # Output: [3, 2, 0, 1]
print(solution.circularPermutation(3, 2))  # Output: [2, 6, 7, 5, 4, 0, 1, 3]
131