<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:
    string sortedNumber(int x) {
        int count[10]{};
        while (x > 0) {
            count[x%10]++;
            x /= 10;
        }
        string s;
        for (int i=0; i<10; i++) {
            while (count[i]--) {
                s += char(i + '0');
            }
        }
        return s;
    }
    bool reorderedPowerOf2(int n) {
        if (n == 1 || n == 2 || n == 4 || n == 8) return true;

        string s = sortedNumber(n);
        int p = 4;
        for (int i = 3; i<30; i++) {
            p *= 2;
            string t = sortedNumber(p);
            if (t.size() > s.size()) break; 
            if (s == t) return true;
        }

        return false;
    }
};

auto boom = []{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();
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