Gray Code LeetCode Solution | Easy Approach

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

 Gray Code Solutions

✅Time: O(2n)
✅Space: O(2n)

C++

class Solution {
 public:
  vector<int> grayCode(int n) {
    vector<int> ans{0};

    for (int i = 0; i < n; ++i)
      for (int j = ans.size() - 1; j >= 0; --j)
        ans.push_back(ans[j] | 1 << i);

    return ans;
  }
};

Java

 
class Solution {
  public List<Integer> grayCode(int n) {
    List<Integer> ans = new ArrayList<>();
    ans.add(0);

    for (int i = 0; i < n; ++i)
      for (int j = ans.size() - 1; j >= 0; --j)
        ans.add(ans.get(j) | 1 << i);

    return ans;
  }
}

Python

class Solution:
  def grayCode(self, n: int) -> List[int]:
    ans = [0]

    for i in range(n):
      for j in reversed(range(len(ans))):
        ans.append(ans[j] | 1 << i)

    return ans

Watch Tutorial

Checkout more Solutions here