<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) {
        vector<int> dp(2502, 1);
        int ans = 0;
        // dp[0]=1;
        // ans = max(ans, dp[0]);
        vector<int> prev(1002);

        for (int i=0; i<1002; i++) prev[i]=i;
        int maxind = 0;
        sort(nums.begin(), nums.end());
        for (int i=0; i<nums.size(); i++)
        {
            // dp[i] = dp[i-1];
            for (int j=0 ;j<=i-1; j++)
            {   
                if ((nums[i] % nums[j] == 0 ) && 1+dp[j] > dp[i])
                {
                    dp[i] = 1+dp[j];                    
                    prev[i]=j;
                }
            }

            if (dp[i] > ans) {
                ans = dp[i];
                maxind = i;
		    }
        }
        
        for(int i=0; i<nums.size(); i++) cout<<dp[i]<<" "; cout<<"\n";
        for(int i=0; i<nums.size(); i++) cout<<prev[i]<<" "; cout<<"\n";

        vector<int> res;
        cout<<maxind<<"\n";
        while(prev[maxind] != maxind)
        {
            res.push_back(nums[maxind]);
            maxind = prev[maxind];
        }
        
        res.push_back(nums[maxind]);
        return res;
    }
};
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