<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>

#include <iostream>
class Solution {
public:
    vector<int> count(int n) {
        vector<int> res;
        while (n > 0) {
            res.push_back(n%10);
            n /= 10;

        }
        sort(res.begin(), res.end(), greater<int>()); 
        return res;
    }
    bool reorderedPowerOf2(int n) {
        vector<int> dig_count = count(n);

        int i = 1;
        while (i <= pow(10, 9)){
            if (dig_count == count(i)) {
                return true;
            }
            i = i << 1;
        }
        return false;
    }
};
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