<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:
    unordered_map<int, int> mappings[32];
    unordered_map<int, int> myMap;
    bool reorderedPowerOf2(int n) {
        int power = 0;
        long long val = 1;
        long long limit = INT_MAX;
        limit++;

        while(val < limit)
        {
            int temp = val;
            while(temp)
            {
                mappings[power][temp % 10]++;
                temp /= 10;
            }
            power++;
            val *= 2;
        }

        int temp = n;

        while(temp)
        {
            myMap[temp % 10]++;
            temp /= 10;
        }

        for(int i = 0; i < 32; i++)
        {
            int fine = true;
            for(int digit = 0; digit<10; digit++)
            {
                if(myMap[digit] != mappings[i][digit])
                {
                    fine = false;
                    break;
                }
            }

            if(fine)
                return true;
        }

        return false;
    }
};
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