<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>solve(long long int a){
        vector<int>v;
        while(a!=0){
            v.push_back(a%10);
            a=a/10;
        }
        sort(v.begin(),v.end());
        return v;
    }
    bool reorderedPowerOf2(int n) {
        map<vector<int>,int>mp;
        long long int a=1;
        long long int e=1e9;
        while(a<=e){
            vector<int>tm=solve(a);
            mp[tm]++;
            a=a*2;
        }
        
        vector<int>tm=solve(n);
        if(mp.count(tm)==0)return 0;
        return 1;
    }
};
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