<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());
        vector<int> count(nums.size(),1);
        map<int,int> mp;
        int last=0,mx=INT_MIN;
        for(int i=0;i<nums.size();i++)
        {
            mp[i]=i;
            for(int j=0;j<i;j++)
            {
                if(count[i]<count[j]+1 && nums[i]%nums[j]==0)
                {
                    count[i]=count[j]+1;
                    mp[i]=j;
                }
            }
            if(mx<count[i])
            {
                mx=count[i];
                last=i;
            }

        }
        vector<int> v(count[last]);
        int i=count[last]-1;
        v[i]=nums[last];
        while(mp[last]!=last)
        {
            last=mp[last];
            i--;
            v[i]=nums[last];
            
        }
        return v;
    }
};
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