Problem Description

You need to generate a Gray code sequence of length 2^n for a given integer n.

A Gray code sequence is a special ordering of binary numbers with these properties:

1. Range: All integers in the sequence must be between 0 and 2^n - 1 (inclusive)

2. Starting point: The sequence must begin with 0

3. Uniqueness: Each integer appears exactly once in the sequence

4. Adjacent difference: Any two consecutive numbers in the sequence must differ by exactly one bit in their binary representation

5. Circular property: The first and last numbers in the sequence must also differ by exactly one bit

For example, when n = 2, one valid Gray code sequence is [0, 1, 3, 2]:

• 0 (binary: 00) and 1 (binary: 01) differ by one bit
• 1 (binary: 01) and 3 (binary: 11) differ by one bit
• 3 (binary: 11) and 2 (binary: 10) differ by one bit
• 2 (binary: 10) and 0 (binary: 00) differ by one bit

The solution uses a mathematical formula to convert each integer i from 0 to 2^n - 1 into its corresponding Gray code value using the operation i ^ (i >> 1). This XOR operation with the right-shifted value efficiently generates the Gray code sequence by ensuring each consecutive pair differs by exactly one bit.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The Gray code problem doesn't involve nodes and edges in a graph structure. We're generating a sequence of integers with specific bit-difference properties.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to generate a complete sequence of 2^n integers.

Involves Linked Lists?

• No: The problem doesn't involve linked list data structures. We're working with a sequence of integers.

Does the problem have small constraints?

• Yes: The constraint n is typically small (often n ≤ 16 in practice), which means we're generating at most 2^16 = 65536 numbers. This is manageable for exhaustive exploration approaches.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to generate all valid sequences where each number appears exactly once and adjacent numbers differ by one bit, backtracking is a natural fit. We can build the sequence step by step, backtracking when we can't find a valid next number that hasn't been used and differs by exactly one bit.

Conclusion: The flowchart suggests using a backtracking approach for generating the Gray code sequence. While the provided solution uses a mathematical formula (i ^ (i >> 1)), backtracking would also work by:

1. Starting with 0
2. At each step, trying all numbers that differ by exactly one bit from the current number
3. Backtracking if we hit a dead end (no valid unused numbers available)
4. Ensuring the final number differs by one bit from 0 to complete the cycle

Intuition

While backtracking could solve this problem by exhaustively trying different sequences, there's a beautiful mathematical pattern hidden in Gray codes that leads to a much more elegant solution.

Let's think about what makes Gray codes special: consecutive numbers differ by exactly one bit. If we look at the first few Gray code sequences:

• For n=1: [0, 1] → [00, 01] in binary
• For n=2: [0, 1, 3, 2] → [00, 01, 11, 10] in binary
• For n=3: [0, 1, 3, 2, 6, 7, 5, 4] → [000, 001, 011, 010, 110, 111, 101, 100] in binary

Notice a pattern? When we go from binary number i to its Gray code representation, we're essentially looking at which bits change. The key insight is that a bit in the Gray code is 1 if and only if the corresponding bit in the binary number is different from the bit to its left.

This leads us to the formula: for any binary number, we can get its Gray code by XORing it with itself shifted right by one position. Why does this work?

When we compute i ^ (i >> 1):

• The leftmost bit stays the same (since we're XORing with 0 from the shift)
• Each other bit becomes the XOR of that bit and the bit to its left in the original number
• This ensures that the Gray code bit is 1 only when there's a transition from 0 to 1 or 1 to 0 in the original binary representation

For example, binary 3 is 011:

• 3 >> 1 gives us 001
• 011 ^ 001 = 010 (which is 2 in decimal)
• This is indeed the Gray code for position 3!

This mathematical relationship means we don't need backtracking at all - we can directly compute the Gray code for each position from 0 to 2^n - 1 using this simple formula, giving us an O(2^n) time solution with no recursion needed.

Learn more about Math and Backtracking patterns.

Solution Approach

The solution leverages the direct binary-to-Gray code conversion formula discovered through the mathematical relationship between binary numbers and their Gray code representations.

Core Algorithm: The implementation uses a list comprehension to generate all Gray codes from 0 to 2^n - 1:

[i ^ (i >> 1) for i in range(1 << n)]

Let's break down each component:

1. Range Generation: range(1 << n)

   • 1 << n is equivalent to 2^n, using bit shifting for efficiency
   • This generates all integers from 0 to 2^n - 1

2. Gray Code Conversion: i ^ (i >> 1)

   • For each integer i, we apply the conversion formula
   • i >> 1 shifts the binary representation of i one position to the right
   • ^ performs XOR between i and its right-shifted version
   • This preserves the highest bit and makes each other bit the XOR of adjacent bits in the original number

Why This Works: The formula ensures that consecutive Gray codes differ by exactly one bit. When we increment from i to i+1 in binary, the change pattern in the resulting Gray codes maintains the single-bit difference property.

Example Walkthrough for n=3:

• i=0: 000 ^ 000 = 000 (Gray: 0)
• i=1: 001 ^ 000 = 001 (Gray: 1)
• i=2: 010 ^ 001 = 011 (Gray: 3)
• i=3: 011 ^ 001 = 010 (Gray: 2)
• i=4: 100 ^ 010 = 110 (Gray: 6)
• i=5: 101 ^ 010 = 111 (Gray: 7)
• i=6: 110 ^ 011 = 101 (Gray: 5)
• i=7: 111 ^ 011 = 100 (Gray: 4)

Result: [0, 1, 3, 2, 6, 7, 5, 4] - a valid Gray code sequence!

Time and Space Complexity:

• Time: O(2^n) - we generate exactly 2^n numbers
• Space: O(2^n) - for storing the output list

This approach is optimal since we must generate all 2^n numbers, and each conversion takes O(1) time.

Example Walkthrough

Let's walk through generating a Gray code sequence for n = 2 using the formula i ^ (i >> 1).

Goal: Generate a sequence of length 2² = 4 containing all numbers from 0 to 3, where consecutive numbers differ by exactly one bit.

Step-by-step conversion:

For each position i from 0 to 3, we calculate its Gray code:

Position i = 0:

• Binary: 00
• Right shift by 1: 00 >> 1 = 00
• XOR operation: 00 ^ 00 = 00
• Gray code value: 0

Position i = 1:

• Binary: 01
• Right shift by 1: 01 >> 1 = 00
• XOR operation: 01 ^ 00 = 01
• Gray code value: 1

Position i = 2:

• Binary: 10
• Right shift by 1: 10 >> 1 = 01
• XOR operation: 10 ^ 01 = 11
• Gray code value: 3

Position i = 3:

• Binary: 11
• Right shift by 1: 11 >> 1 = 01
• XOR operation: 11 ^ 01 = 10
• Gray code value: 2

Final sequence: [0, 1, 3, 2]

Verification of Gray code properties:

• 0 (00) → 1 (01): differ in rightmost bit ✓
• 1 (01) → 3 (11): differ in leftmost bit ✓
• 3 (11) → 2 (10): differ in rightmost bit ✓
• 2 (10) → 0 (00): differ in leftmost bit ✓

Each consecutive pair differs by exactly one bit, including the wrap-around from last to first, confirming this is a valid Gray code sequence.

Solution Implementation

Time and Space Complexity

The time complexity is O(2^n), where n is the input integer. This is because the list comprehension iterates through all integers from 0 to 2^n - 1 (since 1 << n equals 2^n), performing a constant-time XOR operation i ^ (i >> 1) for each integer.

The space complexity is O(1) if we exclude the space required for the output list. The algorithm only uses a loop variable i and performs bitwise operations without requiring any additional data structures. However, if we include the output list in the space complexity analysis, it would be O(2^n) since the returned list contains 2^n elements.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Problem Requirements

Many developers initially try to generate Gray codes by manually ensuring each consecutive pair differs by one bit, leading to complex recursive or iterative solutions with bit manipulation checks.

Pitfall Example:

def grayCode(self, n: int) -> List[int]:
    result = [0]
    used = {0}
  
    while len(result) < (1 << n):
        last = result[-1]
        for i in range(n):
            # Try flipping each bit
            candidate = last ^ (1 << i)
            if candidate not in used and candidate < (1 << n):
                result.append(candidate)
                used.add(candidate)
                break
    return result

Issue: This greedy approach might get stuck or fail to find a valid sequence, as not every path leads to a complete Gray code sequence.

Solution: Use the mathematical formula i ^ (i >> 1) which guarantees a valid Gray code sequence.

2. Incorrect Bit Shifting Priority

Forgetting parentheses around the bit shift operation due to operator precedence.

Pitfall Example:

def grayCode(self, n: int) -> List[int]:
    return [i ^ i >> 1 for i in range(1 << n)]  # Wrong!

Issue: Without parentheses, i ^ i >> 1 is evaluated as (i ^ i) >> 1, which always results in 0 shifted right by 1, giving all zeros.

Solution: Always use parentheses to ensure correct order of operations:

return [i ^ (i >> 1) for i in range(1 << n)]

3. Using Power Operator Instead of Bit Shift

While 2**n works correctly, it's less efficient than bit shifting for powers of 2.

Pitfall Example:

def grayCode(self, n: int) -> List[int]:
    return [i ^ (i >> 1) for i in range(2**n)]

Issue: The ** operator is computationally more expensive than bit shifting, especially for larger values of n.

Solution: Use 1 << n for better performance:

return [i ^ (i >> 1) for i in range(1 << n)]

4. Attempting Recursive Mirror Construction

Some try to build Gray codes using the mirror reflection method recursively, which is correct but more complex and less efficient.

Pitfall Example:

def grayCode(self, n: int) -> List[int]:
    if n == 0:
        return [0]
  
    prev_gray = self.grayCode(n - 1)
    result = prev_gray[:]
  
    for i in range(len(prev_gray) - 1, -1, -1):
        result.append(prev_gray[i] | (1 << (n - 1)))
  
    return result

Issue: While this works, it has O(2^n) space complexity for recursion stack and creates multiple intermediate lists, making it less efficient than the direct formula.

Solution: The direct formula approach is simpler, more efficient, and avoids recursion overhead:

return [i ^ (i >> 1) for i in range(1 << n)]

5. Edge Case Handling for n=0

Not considering that when n=0, the sequence should be [0], not an empty list.

Pitfall Example:

def grayCode(self, n: int) -> List[int]:
    if n == 0:
        return []  # Wrong!
    return [i ^ (i >> 1) for i in range(1 << n)]

Issue: For n=0, there's still one valid Gray code: 0. The range should be [0, 2^0 - 1] = [0, 0].

Solution: The formula naturally handles this case correctly:

return [i ^ (i >> 1) for i in range(1 << n)]  # Works for n=0 too!

When n=0, range(1 << 0) = range(1) = [0], and 0 ^ (0 >> 1) = 0.