Minimum Factorization
<p>Given a positive integer num, return <em>the smallest positive integer </em><code>x</code><em> whose multiplication of each digit equals </em><code>num</code>. If there is no answer or the answer is not fit in <strong>32-bit</strong> signed integer, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> num = 48
<strong>Output:</strong> 68
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> num = 15
<strong>Output:</strong> 35
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>
</ul>
Greedy; Math
class Solution:
    def smallestFactorization(self, num: int) -> int:
        if num < 2:
            return num
        ans, mul = 0, 1
        for i in range(9, 1, -1):
            while num % i == 0:
                num //= i
                ans = mul * i + ans
                mul *= 10
        return ans if num < 2 and ans <= 2**31 - 1 else 0
