<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) {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie();

        int n = nums.size();
        sort(begin(nums), end(nums));
        int maxi = 1, lastind = 0;
        vector<int> dp(n, 1), hash(n);

        for (int i = 1; i < n; i++) {
            hash[i] = i;
            for (int prev = 0; prev < i; prev++) {
                if (nums[i] % nums[prev] == 0 && dp[prev] + 1 > dp[i]) {
                    dp[i] = 1 + dp[prev];
                    hash[i] = prev;
                }
            }
            if (dp[i] > maxi) {
                maxi = dp[i];
                lastind = i;
            }
        }
        deque<int> res;
        res.push_back(nums[lastind]);
        while (lastind != hash[lastind]) {
            lastind = hash[lastind];
            res.push_front(nums[lastind]);
        }
        vector<int> ans(res.begin(), res.end());
        return ans;
    }
};
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