<p>Given a set of <strong>distinct</strong> positive integers <code>nums</code>, return the largest subset <code>answer</code> such that every pair <code>(answer[i], answer[j])</code> of elements in this subset satisfies:</p>

<ul>
	<li><code>answer[i] % answer[j] == 0</code>, or</li>
	<li><code>answer[j] % answer[i] == 0</code></li>
</ul>

<p>If there are multiple solutions, return any of them.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [1,2]
<strong>Explanation:</strong> [1,3] is also accepted.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,4,8]
<strong>Output:</strong> [1,2,4,8]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 2 * 10<sup>9</sup></code></li>
	<li>All the integers in <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());

        vector<vector<int>> subsets;
        int largest_idx = 0;
        int size = 0;

        for (int i = 0; i < nums.size(); ++i) {
            vector<int> tmp;
            tmp.push_back(nums[i]);

            vector<int>* prev = nullptr;
            for (int j = 0; j < i; ++j) {
                if (nums[i] % nums[j] == 0 && (!prev || prev->size() < subsets[j].size()))
                    prev = &subsets[j];
            }

            if (prev) {
                vector<int>& prev_ref = *prev;
                for (int x : prev_ref)
                    tmp.push_back(x);
            }

            subsets.push_back(tmp);
            if (size < tmp.size()) {
                size = tmp.size();
                largest_idx = i;
            }
        }

        return subsets[largest_idx];
    }
};

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