<p>You are given an integer <code>n</code>. We reorder the digits in any order (including the original order) such that the leading digit is not zero.</p>

<p>Return <code>true</code> <em>if and only if we can do this so that the resulting number is a power of two</em>.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> true
</pre>

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

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        power_set = set()
        power_set.add(1)
        cur = 2

        while cur <= 10 ** 10:
            power_set.add(cur)
            cur = cur * 2
        
        n_list = list(str(n))
        size = len(n_list)
        used = [False for i in range(len(n_list))]
        cur = []

        def dfs(c_size):
            if c_size == size:
                if cur[0] != "0":
                    temp = int(''.join(cur))
                    if temp in power_set:
                        return True
                return False
            
            for i in range(size):
                if used[i] != True:
                    cur.append(n_list[i])
                    used[i] = True
                    if dfs(c_size+1):
                        return True
                    used[i] = False
                    cur.pop(-1)
            return False

        if dfs(0):
            return True

        return False
16800