<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) {
        int n = nums.size();
        if (n == 0) return {};
        
        sort(nums.begin(), nums.end());
        vector<int> dp(n, 1), hash(n, 0);
        
      
        for (int i = 0; i < n; ++i) {
            hash[i] = i;  
            for (int j = 0; j < i; ++j) {
                if (nums[i] % nums[j] == 0 && dp[j] + 1 > dp[i]) {
                    dp[i] = dp[j] + 1;
                    hash[i] = j;
                }
            }
        }
        
    
        int max_index = max_element(dp.begin(), dp.end()) - dp.begin();
        
        vector<int> result;
        while (hash[max_index] != max_index) {
            result.push_back(nums[max_index]);
            max_index = hash[max_index];
        }
        result.push_back(nums[max_index]);
        
        reverse(result.begin(), result.end());
        return result;
    }
};

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