<p>An <strong>n-bit gray code sequence</strong> is a sequence of <code>2<sup>n</sup></code> integers where:</p>

<ul>
	<li>Every integer is in the <strong>inclusive</strong> range <code>[0, 2<sup>n</sup> - 1]</code>,</li>
	<li>The first integer is <code>0</code>,</li>
	<li>An integer appears <strong>no more than once</strong> in the sequence,</li>
	<li>The binary representation of every pair of <strong>adjacent</strong> integers differs by <strong>exactly one bit</strong>, and</li>
	<li>The binary representation of the <strong>first</strong> and <strong>last</strong> integers differs by <strong>exactly one bit</strong>.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>any valid <strong>n-bit gray code sequence</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> [0,1,3,2]
<strong>Explanation:</strong>
The binary representation of [0,1,3,2] is [00,01,11,10].
- 0<u>0</u> and 0<u>1</u> differ by one bit
- <u>0</u>1 and <u>1</u>1 differ by one bit
- 1<u>1</u> and 1<u>0</u> differ by one bit
- <u>1</u>0 and <u>0</u>0 differ by one bit
[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].
- <u>0</u>0 and <u>1</u>0 differ by one bit
- 1<u>0</u> and 1<u>1</u> differ by one bit
- <u>1</u>1 and <u>0</u>1 differ by one bit
- 0<u>1</u> and 0<u>0</u> differ by one bit
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [0,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 16</code></li>
</ul>

class Solution:
    def grayCode(self, n: int) -> List[int]:
        # 2^n numbers, so bit length n 
        # 0000 0001 0011.. 1000 
        # 000 100 110 010 011 111 101 001 
        # 0 4 6 2 3 7 5 1
        # 00 10 11 01
        # 0000 1000 1100 0100 0110 1110 

        def toBinary(s):
            k = len(s) -1
            ans = 0
            for bit in s:
                ans += int(bit) * (2**k)
                k-=1
            return ans
        

        '''
        s = "000"
        ret = [0]
        termination = 001
        new_s = 100
        '''
        ret = []
        seen = set()
        s = "".join(["0" for i in range(n)])
        ret.append(toBinary(s))
        seen.add(s)
        termination = s[:-1] + "1"
        while True: 
            if s == termination:
                break 
            for i in range(len(s)):
                new_s = s[:i] + str(1-int(s[i])) + s[i+1:]
                if new_s in seen:
                    continue
                else:
                    s = new_s
                    seen.add(s)
                    ret.append(toBinary(s))
                    break
        return ret
        


            


25800