LeetCode Problem | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++
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.
Use “Ctrl+F” To Find Any Questions Answer. & For Mobile User, You Just Need To Click On Three dots In Your Browser & You Will Get A “Find” Option There. Use These Option to Get Any Random Questions Answer.
About LeetCode
LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews.
LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.
LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.
LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:
- Mathematics/Basic Logical Based Questions
- Arrays
- Strings
- Hash Table
- Dynamic Programming
- Stack & Queue
- Trees & Graphs
- Greedy Algorithms
- Breadth-First Search
- Depth-First Search
- Sorting & Searching
- BST (Binary Search Tree)
- Database
- Linked List
- Recursion, etc.
Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.
Link for the Problem – Subsets 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
Aluminium recycling energy efficiency Aluminium scrap catalyst recovery Scrap aluminum alloys
Sustainable metal reclamation Aluminium recycling programs Scrap metal recovery of aluminum
Recycling yard management, Aluminum cable scrap recycling industry, Eco-conscious metal recycling
Scrap metal reclamation Aluminum siding scrap Aluminium sheet recycling
Metal reprocessing and reclaiming, Aluminum electrical cable, Metal scrap utilization
Environmental metal stewardship Aluminium scrap reuse methods Aluminium recycling network
Scrap metal reclaiming and repurposing, Recycling scrap aluminum cables, Scrap metal certificates
Scrap metal repackaging Scrap aluminium database management Aluminium scrap reclaiming
Metal waste refabrication, Sorting aluminum cable scrap, Scrap metal recovery facility
Scrap metal import restrictions Metal scrap market forecasting Iron raw material procurement
Ferrous material quality assurance, Iron scrap reclamation plants, Metal reclaiming plant
Scrap metal recovery yard Ferrous metal recovery and reuse Iron scrap packaging
Ferrous scrap recuperation, Iron scrap compacting, Metal recovery and salvage
Metal waste recycling facilities Ferrous waste management solutions Iron scrap management services
Ferrous material salvage and recovery, Iron scrap remanufacturing, Scrap metal reclamation and reuse
Metal scrap traceability Ferrous waste recovery and recycling Iron waste reclaiming
Ferrous recycling center, Iron waste sorting, Metal waste reduction strategies
Scrap metal inventory Ferrous scrap reclaiming plant Iron waste reclamation and reprocessing
Ferrous metal reclamation, Iron scrap smelting, Scrap metal reuse services