<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) {
        sort(nums.begin(), nums.end());
        vector<vector<int>> memo(nums.size(), vector<int>());
        vector<int> largestSubset{-1, -1, 0};
        for(int i = nums.size() - 1; i >= 0; i--) {
            for(int j = i; j < nums.size(); j++) {
                if(i == j) {
                    memo[i] = {j, -1, 1};
                    continue;
                }

                if(nums[j] % nums[i] == 0) {
                    int last = nums[memo[j][0]];

                    if(last % nums[i] == 0) {
                        if(memo[i][2] < memo[j][2] + 1) {
                            memo[i] = {i, j, memo[j][2] + 1};
                        }
                    }
                }
            }

            if(largestSubset[2] < memo[i][2]) {
                largestSubset = memo[i];
            }
        }
        vector<int> largestSubsetValues;
        while(largestSubset[1] != -1) {
            largestSubsetValues.push_back(nums[largestSubset[0]]);
            largestSubset = memo[largestSubset[1]];
        }

        largestSubsetValues.push_back(nums[largestSubset[0]]);
        return largestSubsetValues;
    }
};
12900