<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:
    void solve(vector<int>& nums,int idx1,int idx2){


    }

    vector<int> largestDivisibleSubset(vector<int>& arr) {
        
        sort(arr.begin(),arr.end());

        vector<int> dp(arr.size(),1);
        vector<int> hash(arr.size(),0);
        int maxi = 1;
        int lastidx = 0;

        for(int i=0;i<arr.size();i++){

            hash[i] = i;
            for(int prev = 0;prev<i;prev++){

                if((arr[prev]%arr[i]==0 || arr[i]%arr[prev]==0 ) && 1+dp[prev]>dp[i]){

                    dp[i] = 1 + dp[prev];
                    hash[i] = prev;
                }
            }
            if(dp[i]>maxi){

                maxi = dp[i];
                lastidx = i;
            }
        }

        vector<int> lis;
        lis.push_back(arr[lastidx]);

        while(hash[lastidx]!=lastidx){

            lastidx = hash[lastidx];
            lis.push_back(arr[lastidx]);
        }

        return lis;


    }
};
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