<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) {
        set<map<int, int>> twoPowerDigits;

        auto getDigits = [](int twoPower) {
            map<int, int> digits;
            while(twoPower) {
                digits[twoPower%10]++;
                twoPower /= 10;
            }
            return digits;
        };
        for(int i = 1, k = 1; i < 31; ++i) {
            twoPowerDigits.insert(getDigits(k));
            k <<= 1;
        }

        return (twoPowerDigits.count(getDigits(n)));
    }
};
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