368. Largest Divisible Subset

Problem Statement

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.

Intuition

The fact that 1 4 9 10 16 24 32

1 4 16 32

The time we select 32, If it is a part we just have to check that
It is divisile by 16, Because 32 will be divisible by 4 and 1 

Sice 16 is divisible by them and so on 

So we can sort the array and check for Longest Divisible Subsequence

Exactly similar to LIS printing code

Links

https://leetcode.com/problems/largest-divisible-subset/description/

Video Links

https://www.youtube.com/watch?v=gDuZwBW9VvM&ab_channel=takeUforward

Approach 1:

Trick to Print LIS

C++

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& arr) {
        sort(arr.begin(), arr.end());
        int n = arr.size();
        vector<int> dp(n,1);
        vector<int> hash(n);

        int last_index = 0;
        int maxi = 1;

        for(int index=0; index<n; index++){
            hash[index] = index;

            for(int prev=0; prev<index; prev++){
                if(arr[index] % arr[prev] == 0 and dp[index] < dp[prev]+1 ){
                    dp[index] = dp[prev] + 1;

                    hash[index] = prev;
                }

                if(dp[index] > maxi){
                    maxi = dp[index];
                    last_index = index;
                }
            }
        }

        vector<int> ans;
        ans.push_back(arr[last_index]);

        while(hash[last_index] != last_index){
            last_index = hash[last_index];
            ans.push_back(arr[last_index]);
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++