<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) {
        if(nums.size() ==1)return nums;
        sort(nums.begin(),nums.end());
        vector<pair<int,vector<int>>>dp(nums.size());
        for(int i=0;i<nums.size();i++){
            dp[i].first = 1;
            dp[i].second.push_back(nums[i]);
        }
        int maxLen=0,maxInd=-1;
        for(int i=1;i<nums.size();i++){
            for(int j=0;j<i;j++){
                if(nums[i] % nums[j] == 0 && dp[i].first < (dp[j].first+1)){
                    dp[i].first = dp[j].first+1;
                    dp[i].second = dp[j].second;
                    dp[i].second.push_back(nums[i]);
                    
                }
                if(dp[i].first > maxLen){
                        maxLen = dp[i].first;maxInd = i;
                    }
            }
        }
        return dp[maxInd].second;
    }
};
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