## Subsets II

• Time:O(n \cdot 2^n)
• Space:O(n \cdot 2^n)

## C++

class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<vector<int>> ans;
sort(begin(nums), end(nums));
dfs(nums, 0, {}, ans);
return ans;
}

private:
void dfs(const vector<int>& nums, int s, vector<int>&& path,
vector<vector<int>>& ans) {
ans.push_back(path);

for (int i = s; i < nums.size(); ++i) {
if (i > s && nums[i] == nums[i - 1])
continue;
path.push_back(nums[i]);
dfs(nums, i + 1, move(path), ans);
path.pop_back();
}
}
};


## JAVA

class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
dfs(nums, 0, new ArrayList<>(), ans);
return ans;
}

private void dfs(int[] nums, int s, List<Integer> path, List<List<Integer>> ans) {

for (int i = s; i < nums.length; ++i) {
if (i > s && nums[i] == nums[i - 1])
continue;
dfs(nums, i + 1, path, ans);
path.remove(path.size() - 1);
}
}
}


## Python

class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
ans = []

def dfs(s: int, path: List[int]) -> None:
ans.append(path)
if s == len(nums):
return

for i in range(s, len(nums)):
if i > s and nums[i] == nums[i - 1]:
continue
dfs(i + 1, path + [nums[i]])

nums.sort()
dfs(0, [])
return ans