<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 reorderedPowerOf2(int n) {
        if(n==1)
        return 1;
        map<int,int> mp;
        int c=0;
        while(n){
            c++;
            ++mp[n%10];
            n/=10;
        }
        long long t=2;
        int y=1;
        while(y<=c){
            y=0;
            map<int,int> tt;
            long long u=t;
         while(u){
            y++;
            ++tt[u%10];
            u/=10;
        }  
        bool f=1;
         for(auto it:mp){
            if(it.second!=tt[it.first])
            {
                f=0;
                break;
            }
         }
           for(auto it:tt){
            if(it.second!=mp[it.first])
            {
                f=0;
                break;
            }
         }
         cout<<y<<" ";
       if(f)
       return 1;
t*=(long long)2;
        }

return 0;


    }
};
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