Subset-II

Dec 18, 2024 • 5 mins read

Problem Overview

The LeetCode problem titled "Subsets II" focuses on generating all possible subsets of a given set of integers, ensuring that each subset is unique even if the input set contains duplicates. This problem is particularly interesting as it combines elements of recursion, backtracking, and handling duplicates within a set of numbers.

Problem Statement:
Given an integer array nums that may contain duplicates, your task is to return all possible subsets (the power set). The subset should not contain duplicate subsets, and the solution set must not contain duplicate subsets. Return the solution in any order.

Example:

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

Understanding the Problem

When dealing with subsets, especially with potential duplicate elements, the primary challenge is to ensure that no duplicate subsets are included in the final result. This problem can be broken down into the following steps:

Generate all possible subsets using recursion.
Ensure that each repeatedly generated subset is checked for duplicates.
Return only unique subsets.

Key DSA Concepts and Theory

Recursion and Backtracking

Recursion

Recursion involves solving a problem through function calls that divide the problem into smaller instances. A recursive solution typically includes:

Base Case: A simple case with a known solution that stops the recursion.
Recursive Case: A part of the function that reduces the problem to smaller instances.

Backtracking

Backtracking is a refined approach to recursion where we explore all potential options and abandon those paths that do not satisfy the requirements, thus "backtracking" to try other possibilities.

Set/Sorting for Handling Duplicates

To handle duplicate elements, sorting the input initially helps. Sorting ensures duplicates are adjacent, so an in-order traversal (depth-first) can easily skip over duplicates, ensuring only unique subsets are constructed.

Power Set

The power set of a set is the set of all its subsets, including the empty set and the set itself. Our task here is to compute the power set, considering the constraint of no duplicate subsets.

Solution Approach

Step-by-Step Solution

We will approach this solution using recursion and backtracking:

Sort the Input: Start by sorting the input array nums. Sorting helps to manage duplicates easily.
Recursive Backtracking: Implement a recursive function to explore each subset starting from an empty set.
Use of Backtracking: For each number, have the option to either include it in the current subset or not, while ensuring not to reuse duplicates.
Avoid Duplicates: Use a simple check to skip adding a number if it is the same as the one before it and the previous number was not included in the current recursion path.
Store Results: Keep track of all generated subsets and append only unique results to the solution list.

Code Implementation

C++ Code

#include <vector>
#include <algorithm>

class Solution {
public:
    void backtrack(int start, std::vector<int>& nums, std::vector<int>& current, std::vector<std::vector<int>>& result) {
        result.push_back(current);
        for (int i = start; i < nums.size(); ++i) {
            if (i > start && nums[i] == nums[i - 1]) continue; // skip duplicates
            current.push_back(nums[i]);
            backtrack(i + 1, nums, current, result);
            current.pop_back(); // backtrack
        }
    }

    std::vector<std::vector<int>> subsetsWithDup(std::vector<int>& nums) {
        std::vector<std::vector<int>> result;
        std::sort(nums.begin(), nums.end());
        std::vector<int> current;
        backtrack(0, nums, current, result);
        return result;
    }
};

Java Code

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums); // sort to handle duplicates
        backtrack(0, nums, new ArrayList<>(), result);
        return result;
    }

    private void backtrack(int start, int[] nums, List<Integer> current, List<List<Integer>> result) {
        result.add(new ArrayList<>(current));
        for (int i = start; i < nums.length; ++i) {
            if (i > start && nums[i] == nums[i - 1]) continue; // skip duplicates
            current.add(nums[i]);
            backtrack(i + 1, nums, current, result);
            current.remove(current.size() - 1); // backtrack
        }
    }
}

Time and Space Complexity Analysis

Time Complexity: O(2^n) - Given n elements, there are 2^n possible subsets. Each subset operation is performed in constant time due to pushing and popping elements, thus making the time consumption exponential in terms of subset generation.
Space Complexity: O(n) - The space complexity is governed by the recursion stack depth and storage of elements inside it. This is directly proportional to n, the size of the input nums.

Common Mistakes to Avoid

Not Sorting the Input: Sorting is crucial for efficiently handling duplicates. Failing to sort will result in not correctly identifying duplicates next to each other.
Incorrect Duplication Check: Ensure to check for duplicates only when moving from one recursion level to the next. Duplicate check should respect the current recursion level's changes.

Additional Resources and References

Articles on Recursion and Backtracking: Understanding these concepts is crucial for a vast array of problems related to state exploration.
LeetCode Discussions: Engage with the community or seek insights on variations in problem-solving techniques.
Online Courses: Platforms like Coursera and Udemy offer excellent courses on data structures and algorithms, focusing on recursion and backtracking.

By understanding and mastering the approach detailed above, one can efficiently solve this problem and tackle similar challenges requiring unique subset generation with duplicate handling.

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