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, 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 2ⁿ. 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(2ⁿ), Storage Complexity: O(2ⁿ)

Code

    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