medium Gray Code

Problem Statement

Given a positive integer n, return any valid n-bit gray code sequence.

An n-bit gray code sequence is a sequence containing integers where:

• Each integer is in the inclusive range [0, 2ⁿ - 1].
• Sequence always starts with 0.
• All integers are unique in the sequence.
• The binary representation of every pair of adjacent integers differs by exactly one bit.
• The binary representation of the first and last integers differs by exactly one bit.

Examples

• Example 1

  • Input: n = 4
  • Expected Output: [0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8]
  • Justification: This sequence starts at 0 and each subsequent number differs from the previous one by exactly one bit. The sequence covers all numbers from 0 to 15 (which is 2^4 - 1), adhering to the Gray Code property. It also ensures that the first and last numbers (0 and 8) differ by only one bit.

• Example 2

  • Input: n = 1
  • Expected Output: [0, 1]
  • Justification: With a single bit, the sequence can only be [0, 1], as both numbers differ by exactly one bit, and the sequence starts with 0.

• Example 3

  • Input: n = 3
  • Expected Output: [0, 1, 3, 2, 6, 7, 5, 4]
  • Justification: This sequence for 3 bits starts at 0, and each subsequent number differs from the previous by exactly one bit. Additionally, the first and last numbers (0 and 4) differ by only one bit, satisfying all the conditions of Gray Code.

Solution

The approach to generating a Gray Code sequence for a given bit size, n, exploits the recursive nature of Gray Codes. By starting with a base case for a 0-bit sequence, which is simply [0], we iteratively construct the Gray Code sequence for the next bit size by mirroring the existing sequence and prefixing mirrored parts with 1s.

This method effectively doubles the sequence size at each step, ensuring each number differs by exactly one bit from its predecessor. It's a clever exploitation of binary properties, ensuring that as we progress through bit levels, the essential characteristic of Gray Codes – that each consecutive number differs from the last by exactly one bit – is maintained. This approach is both elegant and efficient, leveraging the mathematical properties of binary numbers to satisfy the problem's constraints.

Step-by-Step Algorithm

1. Initialize the Sequence:

   • Start with a sequence containing a single element, 0. This serves as the base case for n = 0, where n is the number of bits.

2. Iterate Over Bits:

   • For each bit from 1 to n (inclusive), perform the following steps to construct the Gray Code sequence for that bit level.

3. Calculate Prefix Value:

   • For the current bit level i, calculate the prefix value addition as 2^(i-1). This value will be used to generate the mirrored part of the sequence.

4. Mirror and Prefix the Sequence:

   • Iterate over the current sequence in reverse order.
   • For each element, add the addition value calculated in step 3 and append this new value to the sequence. This process mirrors the sequence and prefixes it with 1s in the binary representation, ensuring each consecutive number differs by exactly one bit.

5. Repeat Until n Bits:

   • Repeat steps 2-4 for each bit level until you reach n bits. The final sequence after the last iteration is the desired Gray Code sequence for n bits.

Algorithm Walkthrough

Let's consider the Input: n = 4.

1. Initialize the Sequence:

   • Start with the sequence: [0].

2. Bit Level 1 (i = 1):

   • addition = 2^(1-1) = 1.
   • Mirror and prefix: Original sequence [0], mirrored and prefixed [1].
   • New sequence: [0, 1].

3. Bit Level 2 (i = 2):

   • addition = 2^(2-1) = 2.
   • Original sequence: [0, 1].
   • Mirrored and prefixed (with 2 added): [3, 2] (since 1+2=3 and 0+2=2).
   • New sequence: [0, 1, 3, 2].

4. Bit Level 3 (i = 3):

   • addition = 2^(3-1) = 4.
   • Original sequence: [0, 1, 3, 2].
   • Mirrored and prefixed (with 4 added): [6, 7, 5, 4] (since 2+4=6, 3+4=7, 1+4=5, 0+4=4).
   • New sequence: [0, 1, 3, 2, 6, 7, 5, 4].

5. Bit Level 4 (i = 4):

   • addition = 2^(4-1) = 8.
   • Original sequence: [0, 1, 3, 2, 6, 7, 5, 4].
   • Mirrored and prefixed (with 8 added): Reverse iterate through the sequence and add 8 to each:
     • 4+8=12, 5+8=13, 7+8=15, 6+8=14, 2+8=10, 3+8=11, 1+8=9, 0+8=8.
   • New sequence: [0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8].

Code

import java.util.ArrayList;
import java.util.List;

public class Solution {

  // Generates the Gray Code sequence for a given number of bits, n
  public List<Integer> grayCode(int n) {
    List<Integer> result = new ArrayList<>();
    result.add(0); // Initialize with the base case

    for (int i = 0; i < n; i++) {
      int addition = 1 << i; // Calculate the prefix value for the mirrored numbers
      for (int j = result.size() - 1; j >= 0; j--) {
        // Prefix the mirrored sequence with 1s and append to the original sequence
        result.add(result.get(j) + addition);
      }
    }
    return result;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    System.out.println("Example 1 Output: " + solution.grayCode(4));
    System.out.println("Example 2 Output: " + solution.grayCode(1));
    System.out.println("Example 3 Output: " + solution.grayCode(3));
  }
}

Complexity Analysis

Time Complexity:

The primary reason behind the time complexity being is that for each bit added, the number of sequence elements doubles. This exponential growth is due to the mirroring and appending process which effectively doubles the length of the sequence for each increment in n.

Space Complexity:

Space complexity is also since the algorithm stores a sequence of numbers that grows exponentially with the number of bits n. The sequence length equals , as it includes every possible combination of n bits, thereby necessitating storage proportional to the number of elements in the sequence.