In this post, we are going to solve the Subsets II Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

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
Now, let’s see the leetcode solution of Subsets II Leetcode Solution.
Subsets II Leetcode 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
Subsets II Leetcode Solution in CPP
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 Leetcode 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); } } }
Note: This problem Subsets II is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
NEXT: Decode Ways