<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) {
        unordered_map<int, int> divisors;
        int sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (divisors.find(nums[i]) == divisors.end()) {
                vector<int> temp;
                int tempSum = 0;
                int num = nums[i];
                for (int j = 1; j <= sqrt(num); ++j) {
                    if (num % j == 0) {
                        temp.push_back(j);
                        if (num / j != j) temp.push_back(num / j);
                    }
                }
                if (temp.size() == 4) {
                    for (int j = 0; j < temp.size(); ++j) {
                        tempSum += temp[j];
                    }
                    divisors[num] = tempSum;
                    sum += tempSum;
                } else {
                    divisors[num] = 0;
                }
            } else {
                sum += divisors[nums[i]];
            }
        }
        return sum;
    }
};
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