<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>

#include <climits>

class Solution {
public:
    vector<int> closestDivisors(int num) {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        int n1 = num+1;
        int n2 = num+2;
        int mini=INT_MAX;
        int x1,y1;
        for(int i=1;i<=sqrt(n1);i++){
            if(n1%i==0){
                int a = i;
                int b = n1/i;
                int minDif = abs(a-b);
                if(minDif<mini){
                    mini = minDif;
                    x1=a;
                    y1=b;
                }

            }
        }
        for(int i=1;i<=sqrt(n2);i++){
            if(n2%i==0){
                int a = i;
                int b = n2/i;
                int minDif = abs(a-b);
                if(minDif<mini){
                    mini = minDif;
                    x1=a;
                    y1=b;
                }

            }
        }

        return {x1,y1};
    }
};
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