<p>Given an integer <code>num</code>, find the closest two integers in absolute difference whose product equals&nbsp;<code>num + 1</code>&nbsp;or <code>num + 2</code>.</p>

<p>Return the two integers in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = 8
<strong>Output:</strong> [3,3]
<strong>Explanation:</strong> For num + 1 = 9, the closest divisors are 3 &amp; 3, for num + 2 = 10, the closest divisors are 2 &amp; 5, hence 3 &amp; 3 is chosen.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = 123
<strong>Output:</strong> [5,25]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = 999
<strong>Output:</strong> [40,25]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num &lt;= 10^9</code></li>
</ul>

import math
class Solution:
    def closestDivisors(self, num: int) -> List[int]:

        def find_closest_divisors(n):
            divisor = int(math.sqrt(n))
            if divisor*divisor == n:
                return [divisor, divisor]
            
            curr = 1
            result = [-math.inf, math.inf]
            while curr <= divisor:
                if n % curr == 0:
                    if abs(curr - n // curr) < abs(result[0] - result[1]):
                        result[0], result[1] = curr, n // curr
                curr+=1
            
            return result
        

        n1 = find_closest_divisors(num + 1)
        n2 = find_closest_divisors(num + 2)

        if abs(n1[0] - n1[1]) < abs(n2[0] - n2[1]):
            return n1
        else:
            return n2
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