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
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