<p>Given an integer array <code>nums</code>, return <em>the sum of divisors of the integers in that array that have exactly four divisors</em>. If there is no such integer in the array, return <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [21,4,7]
<strong>Output:</strong> 32
<strong>Explanation:</strong> 
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.
</pre>

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

<pre>
<strong>Input:</strong> nums = [21,21]
<strong>Output:</strong> 64
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int sumFourDivisors(vector<int>& nums) {


        // 100
        // 1 100
        // 2 50

        // 50 / 2 = 25


        int res = 0;
        for (int i = 0; i < nums.size(); i++) {

            vector<int> vals;
            int potentialDivisor = 2;

            while (potentialDivisor * potentialDivisor <= nums[i]) {

                if (nums[i] % potentialDivisor == 0) {

                    vals.push_back(potentialDivisor);

                    if (nums[i] / potentialDivisor != potentialDivisor) {
                        vals.push_back(nums[i] / potentialDivisor);
                    }
                }
                potentialDivisor++;
            }

            if (vals.size() == 2) {

                for (auto val: vals) {
                    res += val;
                }
                res += nums[i] + 1;
            }

        }

        return res;
        
    }
};
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