<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) {
        vector<long long> v;
        long long i = 1;
        v.push_back(1);
        while (i*2<=1e9+7) {
            v.push_back(i*2);
            i*=2;
        }
        map<int,int> m;
        int temp = n;
        while (temp) {
            //cout<<temp%10;
            m[temp%10]++;
            temp/=10;
        }
        for (int i = 0; i < v.size(); i++) {
            int temp2 = v[i];
            map<int,int> m2;
            while (temp2) {
                m2[temp2%10]++; temp2/=10;
            }
            if (m.size()!=m2.size()) continue;
            bool f = false;
            for (auto e:m) {
                if (m2[e.first]!=e.second) {
                    f=true;break;
                }
            }
            if (!f) return true;
        }
        return false;
    }
};
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