Largest Divisible Subset

Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies:

Si % Sj = 0 or Sj % Si = 0.

If there are multiple solutions, return any subset is fine.

Example 1:

Input: [1,2,3]
Output: [1,2] (of course, [1,3] will also be ok)

Example 2:

Input: [1,2,4,8]
Output: [1,2,4,8]

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        // Sorting helps us reduce condition to only 1
        // i.e -> nums[i]%nums[j]==0, where i>j
        Arrays.sort(nums);
        List<Integer>[] dp = new ArrayList[nums.length];
        for (int i = 0; i < nums.length; i++)
            dp[i] = new ArrayList<>();

        int max = 0;
        List<Integer> ans = new ArrayList<>();
        // O(N*2N) => O(N^2)
        for (int i = 0; i < nums.length; i++) {
            // O(N)
            for (int j = 0; j < i; j++)
                if (nums[i] % nums[j] == 0 && dp[j].size() + 1 > dp[i].size())
                    // Copying by reference O(1)
                    dp[i] = dp[j];
            // Copying by value O(N)
            // Instead of copying every loop, we can just copy once in the end
            // This will help us reduce time complexity from O(N^3) to O(N^2)
            List<Integer> temp = new ArrayList<>();
            for (int x : dp[i])
                temp.add(x);
            temp.add(nums[i]);
            dp[i] = temp;
            if (dp[i].size() > max) {
                max = dp[i].size();
                ans = dp[i];
            }
        }
        return ans;
    }
    // O(n) space Solution
    public List<Integer> largestDivisibleSubset(int[] nums) {
        int n = nums.length;
        int[] count = new int[n];
        int[] pre = new int[n];
        Arrays.sort(nums);
        int max = 0, index = -1;
        for (int i = 0; i < n; i++) {
            count[i] = 1;
            pre[i] = -1;
            for (int j = i - 1; j >= 0; j--) {
                if (nums[i] % nums[j] == 0) {
                    if (1 + count[j] > count[i]) {
                        count[i] = count[j] + 1;
                        pre[i] = j;
                    }
                }
            }
            if (count[i] > max) {
                max = count[i];
                index = i;
            }
        }
        List<Integer> res = new ArrayList<>();
        while (index != -1) {
            res.add(nums[index]);
            index = pre[index];
        }
        return res;
    }
}