# Solving Leetcode Interviews in Seconds with AI: Circular Permutation in Binary Representation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1238" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given 2 integers n and start. Your task is return any permutation p of (0,1,2.....,2^n -1) such that :  p[0] = start p[i] and p[i+1] differ by only one bit in their binary representation. p[0] and p[2^n -1] must also differ by only one bit in their binary representation.    Example 1: 
 Input: n = 2, start = 3
 Output: [3,2,0,1]
 Explanation: 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]
  Example 2: 
 Input: n = 3, start = 2
 Output: [2,6,7,5,4,0,1,3]
 Explanation: The binary representation of the permutation is (010,110,111,101,100,000,001,011).
    Constraints:  1 <= n <= 16 0 <= start < 2 ^ n 

	# Explanation
	Here's the breakdown:

*   **Gray Code Generation:** The core idea is to generate a Gray code sequence of length 2<sup>n</sup>. Gray code ensures that consecutive numbers differ by only one bit.
*   **Rotation/Offset:** We need to rotate the Gray code sequence such that it starts with the given `start` value. This is achieved by XORing each Gray code element with the `start` value. This maintains the one-bit difference property.

*   **Runtime Complexity:** O(2<sup>n</sup>), **Storage Complexity:** O(2<sup>n</sup>)

	
	# Code
	```python
	def circular_permutation(n: int, start: int) -> list[int]:
    """
    Generates a circular permutation of integers from 0 to 2^n - 1 such that:
    - p[0] = start
    - p[i] and p[i+1] differ by only one bit in their binary representation.
    - p[0] and p[2^n -1] also differ by only one bit in their binary representation.
    """
    result = []
    for i in range(2**n):
        gray_code = i ^ (i >> 1)  # Generate Gray code
        result.append(gray_code ^ start)  # XOR with start to rotate sequence

    return result
	```