368 Largest Divisible Subset

368. 1. Question

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:

nums: [1,2,3]

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

Example 2:

nums: [1,2,4,8]

Result: [1,2,4,8]

2. Implementation

思路: 因为要找的是最大的subset，我们发现最后我们要找的结果和它当前的位置是无关的，所以为了方便比较，我们先对输入的数组排序。然后我们用类似于找Longest Increasing Subsequence的方式找到最大的divisible subset。我们用count和preIndex两个数组分别记录以下信息: count[i]表示在位置i之前的，最大的divisible subset的size，preIndex[i]记录i之前的上一个divisible subset的元素。比如如果 nums[i] % nums[j] == 0 and j < i, preIndex[i] = j. 最后有一点要注意的是，如果subset只有1个元素，那么很明显这是个合法的解(自己余自己等于0)，所以第一个for loop里我们先从i = 0开始，而不是i = 1

(1) DP

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        List<Integer> res = new ArrayList<>();

        if (nums == null || nums.length == 0) {
            return res;
        }

        Arrays.sort(nums);
        int n = nums.length;
        int[] count = new int[n];
        int[] preIndex = new int[n];
        int max = 0, index = -1;


        for (int i = 0; i < n; i++) {
            count[i] = 1;
            preIndex[i] = -1;
            for (int j = i - 1; j >= 0; j--) {
                if (nums[i] % nums[j] == 0 && (1 + count[j] > count[i])) {
                    count[i] = count[j] + 1;
                    preIndex[i] = j;
                }
            }
             if (count[i] > max) {
                max = count[i];
                index = i;
            }
        }

        while (index != -1) {
            res.add(nums[index]);
            index = preIndex[index];
        }
        return res;
    }
}

3. Time & Space Complexity

DP: 时间复杂度O(n^2), 空间复杂度O(n)

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