<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>res;
    void helper(int ind,vector<int>&nums,vector<int>r)
    {
        if(ind==nums.size())
        {
            if(r.size()>res.size())
                res=r;
            return;
        }
        // not piking
        helper(ind+1,nums,r);

        // picking
        if(r.size()==0 || nums[ind]%r.back()==0)
        {
            r.push_back(nums[ind]);
            helper(ind+1,nums,r);
        }
    }
    vector<int> largestDivisibleSubset(vector<int>& nums)
    {
        vector<int>r;
        sort(nums.begin(),nums.end());
        vector<vector<int>>dp(nums.size());
        vector<int>res;
        for(int i=0;i<nums.size();i++)
        {
            for(int j=0;j<i;j++)
            {
                if(nums[i]%nums[j]==0)
                {
                    if(dp[i].size()<dp[j].size()+1)
                    {
                        dp[i]=dp[j];
                    }
                }
            }
            dp[i].push_back(nums[i]);
            if(dp[i].size()>res.size())
                res=dp[i];
        }
        return res;
        
    }
};
15100