Gray Code

Medium

Asked by:

30 views

Topics:

Bit Manipulation

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum value of n? Are there any limitations on the size of the input?
Are we looking for any specific Gray code sequence, or can I return any valid one?
Should the output list be sorted in any specific order, beyond the Gray code property of adjacent elements differing by one bit?
Can n be zero? If so, what should the expected output be?
Are there any memory constraints I should be aware of?

Brute Force Solution

Approach

The brute force method for Gray Code generation means trying every possible sequence of binary numbers. We'll build sequences and check if they meet the Gray Code criteria: that adjacent numbers differ by only one bit.

Here's how the algorithm would work step-by-step:

Start with a list containing only '0'.
For the desired number of bits, repeat the following steps:
Create a reversed copy of the current list.
Add a '0' to the beginning of each number in the original list.
Add a '1' to the beginning of each number in the reversed list.
Combine the modified original list and the modified reversed list to form the new list.
At the end, the list will contain the Gray Code sequence.

Code Implementation

def gray_code_brute_force(number_of_bits):
    gray_code_list = ['0']

    # Iterate to generate Gray code for the specified number of bits
    for _ in range(number_of_bits):
        reversed_list = gray_code_list[::-1]

        # Prepend '0' to original list and '1' to reversed list

        original_list_with_zero = ['0' + code for code in gray_code_list]

        reversed_list_with_one = ['1' + code for code in reversed_list]

        # Combine the modified lists to form the next Gray code sequence
        gray_code_list = original_list_with_zero + reversed_list_with_one

    return gray_code_list

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm iterates n times, where n is the desired number of bits. In each iteration, the list of Gray codes doubles in size. Specifically, we reverse the existing list, prepend '0' to the original, prepend '1' to the reversed list, and concatenate. The number of elements in the list grows exponentially with n, reaching 2^n at the end. The reversal and prepending operations each take O(2^n) time in each iteration, dominated by the size of the growing list. Thus the overall time complexity is O(2^n).

Space Complexity

O(2^N) – The algorithm constructs a list of Gray codes iteratively. In each iteration, it creates a reversed copy of the current list and then concatenates modified versions of the original and reversed lists. Since the list doubles in size with each iteration, and this process repeats N times (where N is the desired number of bits), the final list will contain 2^N elements (strings). Thus, the auxiliary space required to store this list grows exponentially with N, leading to a space complexity of O(2^N).

Optimal Solution

Approach

The Gray code problem asks us to create a list of numbers where only one bit changes between adjacent numbers. The most efficient solution leverages the reflective property of Gray codes to build the sequence in a clever way, avoiding brute-force generation.

Here's how the algorithm would work step-by-step:

Start with a list containing just the number zero.
For each bit position (from least significant to most significant), do the following steps:
Take the current list and create its mirrored (reversed) version.
Add '1' at the current bit position (that is, 2 to the power of the current bit position) to all the numbers in the mirrored list.
Append the modified (mirrored) list to the original list. This creates a new, longer list.
Repeat this process for each bit position. The final list is the Gray code sequence.

Code Implementation

def gray_code(number):
    gray_code_list = [0]

    for bit_position in range(number):
        # Create reversed list for reflection.
        mirrored_list = gray_code_list[::-1]

        # Add 2**bit to mirrored list elements
        for index in range(len(mirrored_list)):
            mirrored_list[index] += 2**bit_position

        # Append the modified list to the original
        gray_code_list.extend(mirrored_list)

    return gray_code_list

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm iterates n times, where n is the number of bits. In each iteration, the current list of Gray codes is mirrored, and 2 raised to the current bit position is added to each element in the mirrored list. The size of the Gray code list doubles in each iteration. Starting with a list of size 1, after n iterations, the list contains 2^n elements. Therefore, the time complexity is dominated by the creation and modification of this list which grows exponentially with n, resulting in a time complexity of O(2^n).

Space Complexity

O(2^n) – The algorithm iteratively builds the Gray code sequence. In each iteration, it creates a reversed (mirrored) copy of the current list and appends it to the original list. The list doubles in size with each bit position processed. Since there are 'n' bit positions, the final list will have a size of 2^n. Thus, the space required to store this list is proportional to 2^n, giving us O(2^n) space complexity.

Edge Cases

Case	How to Handle
n = 0	Return a list containing only 0 because 2^0 = 1 and the sequence should start with 0.
n = 1	Return a list containing [0, 1] as this is the simplest non-trivial Gray code.
Large n (e.g., n > 20): Integer Overflow	Ensure the chosen language and data types (e.g., `long`) can handle numbers up to 2^n without integer overflow, impacting the correctness of bitwise operations and list generation.
Negative n	Throw an IllegalArgumentException or similar error as n must be a non-negative integer to represent the number of bits.
n is a very large number close to system limits	Monitor for out-of-memory errors when creating the list of size 2^n, and consider alternative, memory-efficient algorithms if necessary, such as iterative generation.
Multiple possible solutions	The problem statement allows any valid Gray code sequence, so the solution only needs to produce one such sequence, satisfying the one-bit difference constraint.
Bitwise operations for large n values	Confirm bitwise shift operations are behaving correctly and are not losing significant bits when n is large (consider unsigned integers).
Recursive implementation reaching maximum stack depth (large n)	If a recursive solution is used, ensure it's tail-recursive or rewrite it iteratively to avoid stack overflow for large values of n.

n = 0

How to Handle:

Return a list containing only 0 because 2^0 = 1 and the sequence should start with 0.

n = 1

How to Handle:

Return a list containing [0, 1] as this is the simplest non-trivial Gray code.

Large n (e.g., n > 20): Integer Overflow

How to Handle:

Ensure the chosen language and data types (e.g., `long`) can handle numbers up to 2^n without integer overflow, impacting the correctness of bitwise operations and list generation.

Negative n

How to Handle:

Throw an IllegalArgumentException or similar error as n must be a non-negative integer to represent the number of bits.

n is a very large number close to system limits

How to Handle:

Monitor for out-of-memory errors when creating the list of size 2^n, and consider alternative, memory-efficient algorithms if necessary, such as iterative generation.

Multiple possible solutions

How to Handle:

The problem statement allows any valid Gray code sequence, so the solution only needs to produce one such sequence, satisfying the one-bit difference constraint.

Bitwise operations for large n values

How to Handle:

Confirm bitwise shift operations are behaving correctly and are not losing significant bits when n is large (consider unsigned integers).

Recursive implementation reaching maximum stack depth (large n)

How to Handle:

If a recursive solution is used, ensure it's tail-recursive or rewrite it iteratively to avoid stack overflow for large values of n.