Largest Divisible Subset
<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>
Array; Math; Dynamic Programming; Sorting
class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int[] f = new int[n];
        Arrays.fill(f, 1);
        int k = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] % nums[j] == 0) {
                    f[i] = Math.max(f[i], f[j] + 1);
                }
            }
            if (f[k] < f[i]) {
                k = i;
            }
        }
        int m = f[k];
        List<Integer> ans = new ArrayList<>();
        for (int i = k; m > 0; --i) {
            if (nums[k] % nums[i] == 0 && f[i] == m) {
                ans.add(nums[i]);
                k = i;
                --m;
            }
        }
        return ans;
    }
}