<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 {
public:
    vector<int> closestDivisors(int num) {
        int targets[] = {num + 1, num + 2};
        
        int minDifference = INT_MAX;
        vector<int> closestPair;

        for (int t : targets) {
            for (int i = 1; i <= sqrt(t); i++) {
                if (t % i == 0) {
                    int a = i;
                    int b = t / i;
                    int difference = abs(a - b);
                    if (difference < minDifference) {
                        minDifference = difference;
                        closestPair = {a, b};
                    }
                }
            }
        }
        return closestPair;
    }
};
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