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 {
    public boolean reorderedPowerOf2(int n) {
        String target = f(n);
        for (int i = 1; i <= 1000000000; i <<= 1) {
            if (target.equals(f(i))) {
                return true;
            }
        }
        return false;
    }

    private String f(int x) {
        char[] cnt = new char[10];
        for (; x > 0; x /= 10) {
            cnt[x % 10]++;
        }
        return new String(cnt);
    }
}
