Subsets II LeetCode Solution | Easy Approach

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

Input: nums = [1,2,2]
Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]

Example 2:

Input: nums = [0]
Output: [[],[0]]

Constraints:

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10

 Subsets II Solutions

✅Time:O(n2n)
✅Space:O(n2n) 

C++

 Will be updated Soonclass 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) {
    ans.add(new ArrayList<>(path));

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

Python

  Will be updated Soonclass 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

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