<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) {
        //every number (>1) already has 1 and num as its divisors 
        unordered_set<int> divisors;
        int sumDivisors = 0;

        for (int& num : nums) {
            if (num > 1) {
                int numDivisors = 2;
                vector <int> currentDivisors;
                //find numbers with exactly 2 more divisors
                for (int i = 2; i <= sqrt(num); ++i) {
                    if (num % i == 0) {
                        int compliment = num / i;
                        if (compliment == i) {
                            //only add one of them
                            currentDivisors.push_back(compliment);
                            numDivisors++;
                            continue;
                        }
                        //add both of them if they are unique
                        currentDivisors.push_back(i);
                        currentDivisors.push_back(compliment);
                        numDivisors+= 2;
                    }
                }
                //add them to the sum
                if (numDivisors == 4) {
                    sumDivisors += 1;
                    sumDivisors += num;
                    for (int integer : currentDivisors) {
                        sumDivisors += integer;
                    }
                }

            }
        }

        return sumDivisors;
    }
};
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