# LeetCode 90. Subsets II

## Description

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set). Note: The solution set must not contain duplicate subsets. Example: Input: [1,2,2] Output: [ , , [1,2,2], [2,2], [1,2], [] ]

## Explanation

The problem is a follow-up question for Subset. It is also a typical backtracking coding problem.

We can have a recursion function to add visited subsets to the final results. Remember to make a deep copy whenever we are adding the subset to the results.

## Java Solution

``````class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> results = new ArrayList<>();

if (nums == null || nums.length == 0) {
return results;
}

Arrays.sort(nums);

List<Integer> subset = new ArrayList<>();
toFindAllSubsets(nums, results, subset, 0);

return results;
}

private void toFindAllSubsets(int[] nums, List<List<Integer>> results, List<Integer> subset, int startIndex) {

for (int i = startIndex; i < nums.length; i++) {
if (i != startIndex && nums[i] == nums[i - 1]) {
continue;
}

toFindAllSubsets(nums, results, subset, i + 1);
subset.remove(subset.size() - 1);
}
}
}``````

## Python Solution

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

self.helper(results, [], 0, nums)

return results

def helper(self, results, combination, start, nums):
if combination not in results:
results.append(list(combination))

for i in range(start, len(nums)):
combination.append(nums[i])
self.helper(results, combination, i + 1, nums)
combination.pop()

``````
• Time Complexity: ~N
• Space Complexity: ~N