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# Gray Code LeetCode Solution

## Problem – Gray Code LeetCode Solution

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 LeetCode Solution in Python

``````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
``````

## Gray Code LeetCode Solution in C++

``````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;
}
};
``````

## Gray Code LeetCode Solution in Java

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