# Leetcode Subset problem solution

In this Leetcode Subset problem solution we have Given an integer array nums of unique elements, return all possible subsets (the power set). The solution set must not contain duplicate subsets. Return the solution in any order.

## Problem solution in Python.

```class Solution:
def _subsets(self, nums:List[int], current: List[int], index:int):
if index>=len(nums):
self.sol.append(current)
return
self._subsets(nums, current+[nums[index]], index+1)
# solution ignoring nums[index]
self._subsets(nums, current, index+1)

def subsets(self, nums: List[int]) -> List[List[int]]:
self.sol = []
self._subsets(nums,[], 0)
return self.sol
```

## Problem solution in Java.

```class Solution {
public List<List<Integer>> subsets(int[] nums) {

List<List<Integer>> res = new ArrayList<>();
boolean[] used = new boolean[nums.length];
rec(res, new ArrayList<>(), nums, 0, used);

return res;
}

private void rec(List<List<Integer>> res, List<Integer> temp, int[] nums, int idx, boolean[] used){

if(temp.size() == nums.length){
return;
}

for(int i = idx; i < nums.length; i++){
if(used[i]){
continue;
}
used[i] = true;
rec(res, temp, nums, i, used);
used[i] = false;
temp.remove(temp.size() - 1);
}
}
}
```

## Problem solution in C++.

```class Solution {
public:

vector<vector<int>> ans;
void check(vector<int> nums,int start,vector<int> &v)
{
ans.push_back(v);
for(int i=start;i<nums.size();i++)
{
v.push_back(nums[i]);
check(nums,i+1,v);
v.pop_back();
}
}
vector<vector<int>> subsets(vector<int>& nums) {
vector<int> v;
check(nums,0,v);
return ans;
}
};
```

## Problem solution in C.

```int** subsets(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
if (nums == NULL || numsSize == 0)
return NULL;
int outputSize = pow(2, numsSize);
*returnSize = outputSize;
int **result = (int **)malloc(outputSize * sizeof(int*));
for (int i = 0; i < outputSize; i++)
result[i] = (int *)malloc(numsSize * sizeof(int*));
*returnColumnSizes = (int*)calloc(*returnSize, sizeof(int));

for (int i = 0; i < outputSize; i++)
{
int cnt = 0;
for (int j = 0; j < numsSize; j++)
{
int element = pow(2, j);
if (element & i)
{
result[i][cnt] = nums[j];
cnt++;
}
}
(*returnColumnSizes)[i] = cnt;
}
return result;
}
```