<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:

    multiset<int> get_digits (int n) {
        multiset<int> map;
        while (n > 0) {
            int d = n % 10;
            map.insert(d);
            n /= 10;
        }

        return map;
    }

    bool is_digits_match(multiset<int> a, multiset<int> b) {
        for (int d = 0; d < 10; d++) {
            if (a.count(d) != b.count(d)) {
                return false;
            }
        }
        return true;
    }

    bool reorderedPowerOf2(int n) {

        multiset<int> digits_of_n = get_digits(n);

        for (int i = 0, pow_of_2 = 1; i <= 30; i++ ) {
            multiset<int> digits_of_pow_of_2 = get_digits(pow_of_2);
            pow_of_2 <<= 1;
            if (is_digits_match(digits_of_n, digits_of_pow_of_2)) {
                printf("pow_of_2: %d\n", pow_of_2);
                return true;
            }
        }

        return false;
    }
};
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