An n-bit gray code sequence is a sequence of 2^n integers where:

• Every integer is in the inclusive range [0, 2^n - 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

In this question, we are dealing with binary representation of integers, we will be using bit manipulation. To learn more, checkout Bitmask Intro.

The two important operations we will use are the shift operator (<<) and the XOR (exclusive or) operator (^).

• 1 << i: will shift 1 by i bits to the left, the new binary number will have i zeros after a 1.
• code ^ (1 << i): flips the ith bit (from the right) in code's binary representation. By the property of XOR that 0^1 = 1 and 1^1 = 0.

Now, we can use these two operations to find the candidates for the next code. And since we are only looking for one solution, we can use the aggregation template to return earlier. We will apply the backtracking2 template and fill in the logic.

• is_leaf: start_index == 2**n, we have used all integers
• get_edges: new_code = code ^ (1 << i) where 0 <= i < n, this new_code has one flipped bit on the i-th bit, it can potentially be the next code
• is_valid: new_code is valid when we have not used it (not in visited)

Implementation

def grayCode(self, n: int) -> List[int]:
    length = 1 << n   # same as 2**n
    visited = [False] * length

    def dfs(start_index, code):
        if start_index == length:
            return True
        for i in range(n):
            new_code = code ^ (1 << i)
            print(new_code)
            if not visited[new_code]:
                path.append(new_code)
                visited[new_code] = True
                if dfs(start_index+1, new_code): return True
                visited[new_code] = False
                path.pop()
        return False

    path = [0]
    visited[0] = True
    dfs(1, 0)
    return path