<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) {
      vector<vector<int>> lis(nums.size(), vector<int>(0));
      sort(nums.begin(), nums.end());

      for(int i=0; i<nums.size(); i++) {
        for(int j=0; j < i; j++) {
          if (nums[i] % nums[j] == 0 && lis[i].size() < lis[j].size() + 1) {
            lis[i] = lis[j];
          }
        }
        lis[i].push_back(nums[i]);
      }
      int max_ind = 0, maxi = lis[0].size();
      for(int i = 1; i<lis.size(); i++) {
        if (maxi < lis[i].size()) {
          maxi = lis[i].size();
          max_ind = i;
        }
      }
      return lis[max_ind];
    }
};
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