<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 {
private:
    vector<int> helper(int n) {
        vector<int> divisors;

        for (int i = 1; i<= sqrt(n); i++) {
            if ( i * i == n) {
                divisors.push_back(i);
                continue;
            }
            if (n % i == 0 ) {
                divisors.push_back(i);
                divisors.push_back(n / i);
            }
        }

        return divisors;
    }
public:
    int sumFourDivisors(vector<int>& nums) {
        vector<vector<int>> sum;
        for (int i = 0; i< nums.size(); i++) {
            vector<int> divisors = helper(nums[i]);
            if (divisors.size() == 4) {
                sum.push_back(divisors);
            }
        }

        int ans = 0;
        for (int i = 0; i<sum.size(); i++) {
            for (int j = 0; j<4; j++) {
                ans += sum[i][j];
            }
        }

        return ans;
    }
};
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