<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> res;
        
        int n = nums.size();
        if(n == 0){
            return res;
        }
        
        int count[n];
        int pre[n];
        
        sort(nums.begin(), nums.end());
        
        int maxV = 0, index = -1;
        
        for (int i = 0; i < n; i++) {
            count[i] = 1;
            pre[i] = -1;
            for (int j = 0; j <= i - 1; j++) {
                if (nums[i] % nums[j] == 0) {
                    if (1 + count[j] > count[i]) {
                        count[i] = count[j] + 1;
                        pre[i] = j;
                    }
                }
            }
            if (count[i] > maxV) {
                maxV = count[i];
                index = i;
            }
        }
        
        
        while (index != -1) {
            res.push_back(nums[index]);
            index = pre[index];
        }
        return res;
    }
};
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