Table of Contents

# Largest Divisible Subset – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

# C++ Largest Divisible Subset LeetCode Solution

```
``````
class Solution {
public:
vector<int> largestDivisibleSubset(vector<int>& nums) {
if(nums.size()==0){
return nums;
}
sort(nums.begin(),nums.end());
int flag=1;
if(nums[0]==1){
flag=0;
}
else{
nums.push_back(1);
flag=1;
}
sort(nums.begin(),nums.end());
int i,j;
int dp[nums.size()][2];
for(i=nums.size()-1;i>=0;i--){
dp[i][0]=0;
dp[i][1]=i;
for(j=i+1;j<nums.size();j++){
if((nums[j]%nums[i])==0){
if(dp[j][0]>dp[i][0]){
dp[i][0]=dp[j][0];
dp[i][1]=j;
}
}
}
dp[i][0]++;
}
i=0;
vector<int> t;
t.push_back(nums[i]);
while(dp[i][1]!=i){
i=dp[i][1];
t.push_back(nums[i]);
}
if(flag==1){
t.erase(t.begin());
}
return t;
}
};
```

# Java Largest Divisible Subset LeetCode Solution

```
``````
class Solution {
// if we sort the array, every element in a divisibleSubset can be divisible by the element just before it.
// for any element k, its largestDivisibleSubset that ends with k can be formed in the following way:
// use element k after any one of the previous elements that is divisble
public List<Integer> largestDivisibleSubset(int[] nums) {
int[] l = new int[nums.length]; // the length of largestDivisibleSubset that ends with element i
int[] prev = new int[nums.length]; // the previous index of element i in the largestDivisibleSubset ends with element i
Arrays.sort(nums);
int max = 0;
int index = -1;
for (int i = 0; i < nums.length; i++){
l[i] = 1;
prev[i] = -1;
for (int j = i - 1; j >= 0; j--){
if (nums[i] % nums[j] == 0 && l[j] + 1 > l[i]){
l[i] = l[j] + 1;
prev[i] = j;
}
}
if (l[i] > max){
max = l[i];
index = i;
}
}
List<Integer> res = new ArrayList<Integer>();
while (index != -1){
res.add(nums[index]);
index = prev[index];
}
return res;
}
}
```

# Python 3 Largest Divisible Subset LeetCode Solution

```
``````
class Solution(object):
def largestDivisibleSubset(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
n = len(nums)
if n <= 1:
return nums
nums.sort()
dp = [(0, 0)] * n
dp[0] = (1, 0)
maxIndex, maxVal = 0, 1
for i in range(1, n):
dp[i] = max((dp[j][0] + 1, j) for j in range(i + 1) if nums[i] % nums[j] is 0)
if dp[i] > maxVal:
maxIndex, maxVal = i, dp[i]
i, lds = maxIndex, [nums[maxIndex]]
while i != dp[i][1]:
i = dp[i][1]
lds.append(nums[i])
return lds
```

Array-1180

String-562

Hash Table-412

Dynamic Programming-390

Math-368

Sorting-264

Greedy-257

Depth-First Search-256

Database-215

Breadth-First Search-200

Tree-195

Binary Search-191

Matrix-176

Binary Tree-160

Two Pointers-151

Bit Manipulation-140

Stack-133

Heap (Priority Queue)-117

Design-116

Graph-108

Simulation-103

Prefix Sum-96

Backtracking-92

Counting-86

Sliding Window-73

Linked List-69

Union Find-66

Ordered Set-48

Monotonic Stack-47

Recursion-43

Trie-41

Binary Search Tree-40

Divide and Conquer-40

Enumeration-39

Bitmask-37

Queue-33

Memoization-32

Topological Sort-31

Geometry-30

Segment Tree-27

Game Theory-24

Hash Function-24

Binary Indexed Tree-21

Interactive-18

Data Stream-17

String Matching-17

Rolling Hash-17

Shortest Path-16

Number Theory-16

Combinatorics-15

Randomized-12

Monotonic Queue-9

Iterator-9

Merge Sort-9

Concurrency-9

Doubly-Linked List-8

Brainteaser-8

Probability and Statistics-7

Quickselect-7

Bucket Sort-6

Suffix Array-6

Minimum Spanning Tree-5

Counting Sort-5

Shell-4

Line Sweep-4

Reservoir Sampling-4

Eulerian Circuit-3

Radix Sort-3

Strongly Connected Componen-t2

Rejection Sampling-2

Biconnected Component-1