<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>

#include <vector>
#include <unordered_map>
#include <algorithm>

class Solution {
public:
    bool reorderedPowerOf2(int n) {
        // Convert the number into a digit frequency map
        auto getDigitCount = [](int num) {
            std::unordered_map<char, int> count;
            std::string str = std::to_string(num);
            for (char c : str) {
                count[c]++;
            }
            return count;
        };
        
        // Get digit count for the given number
        std::unordered_map<char, int> nCount = getDigitCount(n);
        
        // Check all powers of 2 up to the range
        for (int i = 0; i < 31; ++i) {
            int powerOf2 = 1 << i; // 2^i
            if (getDigitCount(powerOf2) == nCount) {
                return true;
            }
        }
        
        return false;
    }
};

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