**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

An **n-bit gray code sequence** is a sequence of `2`

integers where:^{n}

- Every integer is in the
**inclusive**range`[0, 2`

,^{n}- 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`

```
class Solution:
# @return a list of integers
'''
from up to down, then left to right
0 1 11 110
10 111
101
100
start: [0]
i = 0: [0, 1]
i = 1: [0, 1, 3, 2]
i = 2: [0, 1, 3, 2, 6, 7, 5, 4]
'''
def grayCode(self, n):
results = [0]
for i in range(n):
results += [x + pow(2, i) for x in reversed(results)]
return results
```

```
class Solution {
void utils(bitset<32>& bits, vector<int>& result, int k){
if (k==0) {
result.push_back(bits.to_ulong());
}
else {
utils(bits, result, k-1);
bits.flip(k-1);
utils(bits, result, k-1);
}
}
public:
vector<int> grayCode(int n) {
bitset<32> bits;
vector<int> result;
utils(bits, result, n);
return result;
}
};
```

```
public List<Integer> grayCode(int n) {
List<Integer> result = new LinkedList<>();
for (int i = 0; i < 1<<n; i++) result.add(i ^ i>>1);
return result;
}
```

In our experience, we suggest you solve this Gray Code LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Gray Code LeetCode Solution

I hope this Gray Code LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

**More Coding Solutions >>**