<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> ans;
    void rec(vector<int> &nums, vector<int> &k, int ind,int prev, vector<int> &dp){
        if(ind==nums.size()){
            if(k.size()>ans.size()) ans = k;
            return ;
        }
        rec(nums, k, ind+1, prev, dp);
        if(prev==-1 || dp[ind]<k.size() && (nums[ind]%nums[prev]==0 || nums[prev]%nums[ind]==0)){
            dp[ind] = k.size();
            k.push_back(nums[ind]);
            rec(nums, k, ind+1, ind, dp);
            k.pop_back();
        }

    }
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        vector<int> k;
        vector<int> dp(nums.size()+1, -1);
        sort(nums.begin(), nums.end());
        rec(nums, k, 0, -1, dp);
        return ans;
    }
};
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