<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>

from math import sqrt

class Solution:
    def closestDivisors(self, num: int) -> List[int]:
        plus1 = None
        plus2 = None
        sqrt2 = int(sqrt(num+2))

        while (plus1 == None or plus2 == None) and sqrt2 > 0:

            if plus1 == None:
                if (num + 1) % sqrt2 == 0:
                    plus1 = [sqrt2, (num+1)//sqrt2]

            if plus2 == None:
                if (num + 2) % sqrt2 == 0:
                    plus2 = [sqrt2, (num+2)//sqrt2]

            sqrt2 -= 1

        #print(plus1)
        #print(plus2)
        if self.diff(plus1) < self.diff(plus2):
            return plus1
        else:
            return plus2



    def diff(self, num):
        return abs(num[0] - num[1])


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