Reordered Power of 2
<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>
Hash Table; Math; Counting; Enumeration; Sorting
class Solution:
    def reorderedPowerOf2(self, n: int) -> bool:
        def f(x: int) -> List[int]:
            cnt = [0] * 10
            while x:
                x, v = divmod(x, 10)
                cnt[v] += 1
            return cnt

        target = f(n)
        i = 1
        while i <= 10**9:
            if f(i) == target:
                return True
            i <<= 1
        return False
