<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());
        if(nums.size() == 1) {
            return vector<int>({nums[0]});
        } else if(nums.size() == 2) {
            if(nums[1] % nums[0] == 0) {
                return vector<int>({nums[0], nums[1]});
            } else {
                return vector<int>({nums[0]});
            }
        }

        vector<int> opt(nums.size(),1);
        vector<vector<int>> ans(nums.size());

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

        for(int i=0; i<nums.size(); i++) {
            for(int j=0; j<i; j++) {
                if(nums[i] % nums[j] == 0) {
                    if(opt[i] < opt[j]+1) {
                        opt[i] = opt[j]+1;
                        ans[i] = ans[j];
                        ans[i].push_back(nums[i]);
                    }
                }
            }
        }

        int it = 0;
        for (int i = 1; i < nums.size(); i++) {
            if (ans[i].size() > ans[it].size()) 
                it = i;
        }

        return ans[it];
    }
};
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