<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:

    bool func(int num,vector<int> & digits){

        vector<int>temp(10,0);

        while(num!=0){

            int digit=num%10;
            temp[digit]++;
            num=num/10;
        }

        for(int i=0;i<10;i++){
            if(temp[i]!=digits[i]){
                return false;
            }
        }

        return true;
    }

    bool reorderedPowerOf2(int n) {
        
        vector<int>digits(10,0);
        
        if(n==1){
            return true;
        }

        while(n!=0){

            int digit=n%10;
            digits[digit]++;
            n=n/10;
        }

        long num=1;
        int bits=1;
        while(bits<32){

            if(func(num,digits)){
                return true;
            }

            num=num*2;
            bits++;
        }
        return false;
    }
};
8200