<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>

class Solution:
    def closestDivisors(self, num: int) -> List[int]:
        
        # get divisors of num+1 and num+2
        def getDivisor(n):
            # n is a number
            # arr=[]
            ans=[1,n]
            i=1
            min_diff=float("inf")
            while i*i<=n:
                if n%i==0:
                    d1=i
                    d2=n//i
                    if abs(d1-d2)<min_diff:
                        ans=[d1,d2]
                        min_diff=abs(d1-d2)
                    # arr.append(d1)
                    # arr.append(d2)
                i+=1
            return ans

        f11,f12=getDivisor(num+1)
        f21,f22=getDivisor(num+2)
        if abs(f11-f12)<abs(f21-f22):
            return [f11,f12]
        return [f21,f22]

        # a1=getDivisor(num+1)
        # a1.sort()
        # a2=getDivisor(num+2)
        # a2.sort()
        # f11=a1[len(a1)//2-1]
        # f12=a1[len(a1)//2]

        # f21=a2[len(a2)//2-1]
        # f22=a2[len(a2)//2]

        # if abs(f11-f12)<abs(f21-f22):
        #     return [f11,f12]
        # return [f21,f22]
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