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:

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]

✅ Solution: Tabulation + Hashing (Reconstruction)

class Solution {
public:
    vector<int> largestDivisibleSubset(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());  // Sort to ensure divisibility in order

        vector<int> dp(n, 1);     // dp[i] = length of largest subset ending at i
        vector<int> hash(n, 0);   // hash[i] = previous index in the subset

        for(int i = 0; i < n; i++){
            hash[i] = i; // Point to self initially
            for(int prev = 0; prev < i; prev++){
                if(nums[i] % nums[prev] == 0 && dp[i] < dp[prev] + 1){
                    dp[i] = dp[prev] + 1;
                    hash[i] = prev;
                }
            }
        }

        // Find index of max value in dp[]
        int maxLen = -1, lastIndex = -1;
        for(int i = 0; i < n; i++){
            if(dp[i] > maxLen){
                maxLen = dp[i];
                lastIndex = i;
            }
        }

        // Reconstruct the largest divisible subset
        vector<int> subset;
        subset.push_back(nums[lastIndex]);
        while(hash[lastIndex] != lastIndex){
            lastIndex = hash[lastIndex];
            subset.push_back(nums[lastIndex]);
        }

        reverse(subset.begin(), subset.end()); // Optional: return in increasing order
        return subset;
    }
};

📝 How It Works

You need to find the largest subset where for every pair (i, j), either nums[i] % nums[j] == 0 or nums[j] % nums[i] == 0.
Sort the array → guarantees that if nums[i] % nums[j] == 0, then i > j is more likely.
Build dp[i] as the length of the largest divisible subset ending at index i.
Use a hash array to reconstruct the actual subset.

🧩 Key Transition

if(nums[i] % nums[prev] == 0 && dp[i] < dp[prev] + 1){
    dp[i] = dp[prev] + 1;
    hash[i] = prev;
}

⏱️ Time & Space Complexity

Approach	Time	Space
Tabulation + Hash	O(N²) ✅	O(N) ✅

⚠️ Edge Cases

Single element → return the element
All numbers are powers/multiples of each other
No pair divisible → return any one element

💡 Other Approaches

Method	Notes
Brute Force Subsets	❌ Too slow (2^N)
DP + Hash (this one)	✅ Reconstructs actual subset
Greedy	❌ Fails to build full chain

💬

Problem Statement:​

✅ Solution: Tabulation + Hashing (Reconstruction)​

📝 How It Works​

🧩 Key Transition​

⏱️ Time & Space Complexity​

⚠️ Edge Cases​

💡 Other Approaches​

🔁 Related Problems​

Discussion & Doubts