<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 {
public:
    vector<int> frequency(int n){
        vector<int> freq(10);
        while(n > 0){
            freq[n % 10]++;
            n /= 10;
        }
        return freq;
    }
    bool compare(vector<int>& freq1 , vector<int>& freq2){
        for(int i = 0 ; i < 10 ; i++){
            if(freq1[i] != freq2[i]){
                return false;
            }
        }
        return true;
    }
public:
    bool reorderedPowerOf2(int n) {
        vector<int> newfreq = frequency(n);
        for(int i = 0 ; i < 31 ; i++){
            int number = (int)pow(2 , i);
            vector<int> poweroftwo = frequency(number);
            if(compare(poweroftwo , newfreq)){
                return true;
            }
        }
        return false;
    }
};
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