Leetcode 368: Largest Divisible Subset

Given a set of distinct positive integers nums, return the largest subset answer such that every pair (answer[i], answer[j]) of elements in this subset satisfies:

• answer[i] % answer[j] == 0, or
• answer[j] % answer[i] == 0

If there are multiple solutions, return any of them.

Example 1:

Input: nums = [1,2,3]
Output: [1,2]
Explanation: [1,3] is also accepted.

Example 2:

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

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 2 * 109
• All the integers in nums are unique.

Problem Analysis:

For simplicity, we can first sort the array. Then suppose we obtain such a subset, it is a subarray, then for every i, j < subarray size, and i < j, subarray[j] % subarray[i] == 0.

So we can scan the array, record the largest subset, it is the result we wanted.

Solution

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        vector<pair<int, int>> dp(nums.size(), make_pair(-1, -1));
        int length = 0;
        int last = -1;
        for (int i = 0; i < nums.size(); ++i) {
            for (int j = i + 1; j < nums.size(); ++j) {
                if (nums[j] % nums[i] == 0) {
                    if (dp[j].first == -1) {
                        dp[j] = make_pair(2, i);
                        if (dp[j].first > length) {
                            length = dp[j].first;
                            last = j;
                        }
                    } else if (dp[i].first >= dp[j].first) {
                        dp[j] = make_pair(dp[i].first + 1, i);
                        if (dp[j].first > length) {
                            length = dp[j].first;
                            last = j;
                        }
                    }
                }
            }
        }
        
        if (length == 0) return {nums[0]};
        vector<int> res;
        while (last != -1) {
            res.push_back(nums[last]);
            last = dp[last].second;
        }
        
        reverse(res.begin(), res.end());
        return res;
    }
};

time complexity is O(n²)

Space complexity is O(n)