Subsets II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Subsets II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemSubsets II– LeetCode Problem

Subsets II– LeetCode Problem

Problem:

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– LeetCode Solutions
Subsets II Solution in 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();
    }
  }
};
Subsets II Solution in 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);
    }
  }
}
Subsets II Solution in 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

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