<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) {
        
        int n = nums.size(), id, mx=0, tx=1;
        
        vector<int> res;
        
        int dp[1001];
        memset(dp, 1, sizeof(dp));
        
        vector<vector<int>> s(n);
        
        if(n==1) {res.push_back(nums[0]); return res;}
        
        sort(nums.begin(), nums.end());
        mx = 1, id = n-1, tx = 1;
        
        s[n-1] = {nums[n-1]};
        res = s[n-1];
        for(int i=n-2; i>=0; i--)
        {
            s[i].push_back(nums[i]), mx = 1, id = i;
            
            for(int j=i+1; j<n; j++)
                if(nums[j] % nums[i] ==0) 
                {
                    if(1+dp[j] > dp[i]) dp[i] = 1+dp[j], id = j;
                }
            if(id != i) s[i].insert(s[i].end(), s[id].begin(), s[id].end());
            if(s[i].size()> tx) tx = s[i].size(), res = s[i];
        }
        
        return res;
    }
};
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