<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:
    // Function to return the digit frequency map for a given number
    string countDigits(int num) {
        string count(10, '0'); // to store count of digits from 0 to 9
        while (num > 0) {
            count[num % 10]++;
            num /= 10;
        }
        return count;
    }

    bool reorderedPowerOf2(int n) {
        // Get the digit count of the input number
        string original = countDigits(n);
        
        // Check all powers of 2 up to 2^30 (which is within the range of int)
        for (int i = 0; i < 31; i++) {
            int powerOf2 = 1 << i;  // Compute 2^i
            // If the digit frequency matches, return true
            if (countDigits(powerOf2) == original) {
                return true;
            }
        }

        // If no match is found, return false
        return false;
    }
};
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