<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> dp(nums.size(), 1);
        sort(nums.begin(), nums.end());
        int indMax = 0;
        int maxVal = 0;
        for (int i = 1; i != nums.size(); ++i)
        {
            for (int j = 0; j != i; ++j)
            {
                if ((nums[i] % nums[j]) == 0)
                {
                    dp[i] = max(dp[j] + 1, dp[i]);
                    if (dp[i] > maxVal)
                    {
                        maxVal = dp[i];
                        indMax = i;
                    }
                }
            }
        }
        vector<int> answer{nums[indMax]};
        int prev = nums[indMax];
        int prevVal = maxVal;
        for (int i = indMax - 1; i >= 0; --i)
        {
            if ((prev % nums[i]) == 0 && prevVal == (dp[i] + 1))
            {
                answer.push_back(nums[i]);
                prev = nums[i];
                prevVal = dp[i];
            }
        }
        return answer;
    }
};
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