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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:

Examples

Try it yourself

Try solving this question here:

🎯 STRICT STANDOUT — Gray Code (medium)

1. Why / judgment

Binary reflected Gray code: sequence of all n-bit strings with adjacent Hamming distance 1. Formula code(i)=i XOR (i>>1) for i in 0..2^n-1 is O(1) per value. Reflective construction: n-1 list, prefix 0; reverse with prefix 1. When-NOT: need specific start not 0 -> rotate; Hamiltonian path on hypercube same idea.

2. Big-O derivation (K11)

Output size 2^n -> time Theta(2^n) lower bound.
Formula O(1) per entry.
n=2 -> [0,1,3,2]; n=3 -> [0,1,3,2,6,7,5,4] typical.

3. Pattern + when-NOT (K12)

Name: BINARY REFLECTED GRAY CODE

Recognition: list 0..2^n-1 with adjacent single-bit flips.

When-NOT: Arbitrary Gray-like order without full coverage -> different. Minimal change order for combinations -> combinatorial Gray codes.

4. Edge hand-run (K13)

n=1 -> [0,1]
start must be 0 per problem
uniqueness of sequence not required — any valid

5. Interviewer follow-ups (model answers)

Q1. Why i XOR (i>>1) works?
A: Standard Gray encoding; adjacent i differ by 1 implies codes differ one bit.

Q2. Reflect construction intuition?
A: Mirror previous and set high bit — only high bit flips at mirror junction.

Q3. Space?
A: Theta(2^n) for answer list.

6. Short drills

Drill: generate n=3 by formula
Drill: verify Hamming dist 1 for each adjacent pair
Drill: reflect build n=2 from n=1
✅ Solution 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

java
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.

🎯 STRICT STANDOUT — Solution Gray Code

1. Why / judgment

Loop i=0..2^n-1 append i XOR (i>>1). Hand-run n=2: 0,1,3,2. n=3 list 0,1,3,2,6,7,5,4. Reflect alternative: start [0]; for b in bits mirror with | (1<

2. Big-O derivation (K11)

Theta(2^n) time/space
Bit ops O(1) per
n up to 16-ish by constraints

3. Pattern + when-NOT (K12)

Name: GRAY CODE FORMULA GEN

Recognition: n-bit gray sequence.

When-NOT: Need lexicographic specific path — may still use formula.

4. Edge hand-run (K13)

n=0 edge sometimes [0]
check first is 0

5. Interviewer follow-ups (model answers)

Q1. Prove adjacent?
A: i and i+1 Gray codes differ one bit — textbook.

Q2. Reflect vs formula?
A: Same family; formula shorter code.

Q3. Unsigned shift?
A: Logical >> on non-neg fine.

6. Short drills

Drill: n=1,2,3 lists
Drill: hamming each step
Drill: reflect code

🧩 Pattern · Backtracking

Recognize it: Enumerate all subsets / permutations / combinations under constraints → recurse, choose, un-choose.

▶ Visualize this problem (step it, predict each fork)
⛶ Open this problem debugger in explore mode
🤖 Don't fully get this? Learn it with Claude

Stuck on Gray Code? Open Claude, copy a block below, and it'll teach you this exact concept — visually and interactively.

🪜 Hint ladder (no spoilers)

Progressively stronger hints — you still solve it.

I'm working on the problem **Gray Code** (DSA). Give me a HINT LADDER: start with the tiniest nudge, then wait. Only reveal the next, stronger hint when I ask. Do NOT show the full solution unless I type 'show solution'. Keep me doing the thinking. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
🎨 Explain the approach visually

See the technique, not just code.

Explain the optimal approach to **Gray Code** with a VISUAL walkthrough: trace it on a small concrete example using ASCII art / a step-by-step diagram, narrate what changes each step, then give time & space complexity with a one-line derivation. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
🔍 Review my solution

Catch bugs, edge cases, sub-optimality.

I'll paste my solution to **Gray Code**. Review it for correctness, missed edge cases, and time/space complexity, then coach me toward the optimal — don't just rewrite it. Ask me to paste my code now. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.
🔁 Drill the pattern

Lock in recognition with look-alikes.

Give me 2 problems that use the SAME underlying pattern as **Gray Code**. For each, let me attempt first, then review my answer and name the trigger signal that reveals the pattern. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.

📝 My notes

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