Reordered Power of 2
<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>
Hash Table; Math; Counting; Enumeration; Sorting
impl Solution {
    pub fn reordered_power_of2(n: i32) -> bool {
        fn f(mut x: i32) -> [u8; 10] {
            let mut cnt = [0u8; 10];
            while x > 0 {
                cnt[(x % 10) as usize] += 1;
                x /= 10;
            }
            cnt
        }

        let target = f(n);
        let mut i = 1i32;
        while i <= 1_000_000_000 {
            if target == f(i) {
                return true;
            }
            i <<= 1;
        }
        false
    }
}
