Introduction

In this blog post, we will explore how to solve the LeetCode problem "89" 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

An n-bit gray code sequence is a sequence of 2n integers where: Every integer is in the inclusive range [0, 2n - 1], The first integer is 0, An integer appears no more than once in the sequence, The binary representation of every pair of adjacent integers differs by exactly one bit, and The binary representation of the first and last integers differs by exactly one bit. Given an integer n, return any valid n-bit gray code sequence. Example 1: Input: n = 2 Output: [0,1,3,2] Explanation: The binary representation of [0,1,3,2] is [00,01,11,10]. - 00 and 01 differ by one bit - 01 and 11 differ by one bit - 11 and 10 differ by one bit - 10 and 00 differ by one bit [0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01]. - 00 and 10 differ by one bit - 10 and 11 differ by one bit - 11 and 01 differ by one bit - 01 and 00 differ by one bit Example 2: Input: n = 1 Output: [0,1] Constraints: 1 <= n <= 16

Explanation

• Reflective Binary Gray Code: The core idea is to use the reflective property of Gray codes. An n-bit Gray code can be constructed recursively from an (n-1)-bit Gray code.
  • Recursion (or Iteration): Start with the Gray code for n=1 ([0, 1]). For n=2, reflect this sequence ([1, 0]), prepend a '0' to the original sequence and a '1' to the reflected sequence. This gives [00, 01, 11, 10] which is [0, 1, 3, 2] in decimal. Repeat this process.
  • Bit Manipulation: Instead of using string operations or list concatenation to create the sequences, use bitwise left shift and XOR operations to efficiently generate the Gray code sequence.

• Time Complexity: O(2ⁿ), Space Complexity: O(1) (excluding the output list). Iterative solution uses constant extra space.

Code

    class Solution:
    def grayCode(self, n: int) -> list[int]:
        """
        Generates an n-bit Gray code sequence.

        Args:
            n: The number of bits.

        Returns:
            A list of integers representing the Gray code sequence.
        """

        result = [0]  # Start with the initial Gray code (n=0 or n=1)
        for i in range(n):
            mask = 1 << i  # Create a mask for the current bit position

            # Reflect the existing sequence and XOR with the mask
            for j in range(len(result) - 1, -1, -1):
                result.append(result[j] | mask)
        return result