What is the time complexity of Subsets II?

The optimal solution has a time complexity of O(N*2^N).

What is the space complexity of Subsets II?

The space complexity is O(N*2^N).

What data structures are used to solve Subsets II?

This problem uses Array, Backtracking, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Subsets II?

Insight 1: Sorting the input array `nums` is crucial to efficiently identify and skip duplicate subsets during the backtracking process. Insight 2: The backtracking algorithm explores all possible combinations by recursively adding or not adding each element to a subset.

Which companies ask LeetCode 90. Subsets II in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Google and 4 more companies.

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Subsets II - LeetCode 90 Solution

Q: How to solve Subsets II?

The solution utilizes a backtracking algorithm to generate all possible subsets. First, the input array is sorted to easily identify duplicate numbers. The backtracking function recursively explores all combinations, building subsets incrementally. To avoid generating duplicate subsets, it checks if the current element is the same as the previous element in the sorted array and skips it if they are the same and the index `i` is greater than the starting index `start`.

Subsets II - Complete Solution Guide

Subsets II is LeetCode problem 90, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to generate all possible subsets (the power set) of a given integer array `nums`. The catch is that the input array may contain duplicate elements, and the resulting set of subsets must not contain any duplicate subsets. The order of the subsets in the output doesn't matter.

Solution Approach

The solution utilizes a backtracking algorithm to generate all possible subsets. First, the input array is sorted to easily identify duplicate numbers. The backtracking function recursively explores all combinations, building subsets incrementally. To avoid generating duplicate subsets, it checks if the current element is the same as the previous element in the sorted array and skips it if they are the same and the index `i` is greater than the starting index `start`.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order. This allows us to easily identify and skip duplicate elements.
Step 2: Initialize an empty list `res` to store the resulting subsets.
Step 3: Define a recursive backtracking function `backtrack(start, subset)`. `start` indicates the starting index for the current iteration, and `subset` is the current subset being built.
Step 4: Inside `backtrack`, add a copy of the current `subset` to the `res` list. This represents one valid subset.
Step 5: Iterate through the `nums` array, starting from the `start` index. For each element `nums[i]`, check if it's a duplicate by comparing it to the previous element. If `i > start` (meaning it's not the first element being considered at this level of recursion) and `nums[i] == nums[i - 1]`, skip this element to avoid duplicate subsets.
Step 6: If the element is not a duplicate, add it to the current `subset` and recursively call `backtrack(i + 1, subset)` to explore the subsets that include this element.
Step 7: After the recursive call returns, remove the last element from the `subset` (backtrack) to explore the subsets that do not include this element.
Step 8: Call the `backtrack` function initially with `start = 0` and an empty `subset`. This starts the process of generating all possible subsets.
Step 9: Return the `res` list containing all generated subsets.

Key Insights

Insight 1: Sorting the input array `nums` is crucial to efficiently identify and skip duplicate subsets during the backtracking process.
Insight 2: The backtracking algorithm explores all possible combinations by recursively adding or not adding each element to a subset.
Insight 3: To avoid duplicate subsets, a condition is added within the loop to skip over consecutive duplicate elements in the sorted array, ensuring each unique element is considered only once at each level of the recursion.

Complexity Analysis

Time Complexity: O(N*2^N)

Space Complexity: O(N*2^N)

Topics

This problem involves: Array, Backtracking, Bit Manipulation.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Google, Meta, Microsoft, Swiggy, TikTok.

Subsets II | LeetCode 90

Medium

Array Backtracking Bit Manipulation

Time: O(N*2^N)Space: O(N*2^N)

Solution Code

Available in:

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

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);

        List<Integer> subset = new ArrayList<>();
        backtrack(0, subset, nums, res);
        return res;
    }

    private void backtrack(int start, List<Integer> subset, int[] nums, List<List<Integer>> res) {
        res.add(new ArrayList<>(subset));

        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }

            subset.add(nums[i]);
            backtrack(i + 1, subset, nums, res);
            subset.remove(subset.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 90: Subsets II - Detailed explanation and solution approaches.

Topics

Array Backtracking Bit Manipulation

Companies

AdobeAmazonAppleBloombergGoogleMetaMicrosoftSwiggy

Stats

Difficulty:Medium

Languages:4

Companies:9

Problem #:90

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Sort the input array `nums` in ascending order. This allows us to easily identify and skip duplicate elements.
2
Step 2: Initialize an empty list `res` to store the resulting subsets.
3
Step 3: Define a recursive backtracking function `backtrack(start, subset)`. `start` indicates the starting index for the current iteration, and `subset` is the current subset being built.
4
Step 4: Inside `backtrack`, add a copy of the current `subset` to the `res` list. This represents one valid subset.
5
Step 5: Iterate through the `nums` array, starting from the `start` index. For each element `nums[i]`, check if it's a duplicate by comparing it to the previous element. If `i > start` (meaning it's not the first element being considered at this level of recursion) and `nums[i] == nums[i - 1]`, skip this element to avoid duplicate subsets.
6
Step 6: If the element is not a duplicate, add it to the current `subset` and recursively call `backtrack(i + 1, subset)` to explore the subsets that include this element.
7
Step 7: After the recursive call returns, remove the last element from the `subset` (backtrack) to explore the subsets that do not include this element.
8
Step 8: Call the `backtrack` function initially with `start = 0` and an empty `subset`. This starts the process of generating all possible subsets.
9
Step 9: Return the `res` list containing all generated subsets.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N*2^N)

💾

Space Complexity

O(N*2^N)

📝

Why?

The time complexity arises from generating 2^N subsets, and copying each subset into the result. The space complexity is mainly due to storing all the generated subsets which are 2^N in number. The recursion depth is N, which is also accounted in the space complexity, but 2^N dominates N.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Approach with Bit Manipulation

An iterative approach can also be implemented using bit manipulation. Each bit in a number from 0 to 2^N - 1 represents whether or not the corresponding element in `nums` is included in the subset. This approach requires more complex logic to handle duplicates.

Time: O(N*2^N)Space: O(N*2^N)

Trade-offs: While conceptually interesting, the bit manipulation approach is often harder to understand and implement correctly, especially when dealing with duplicates. The backtracking approach is generally cleaner and more intuitive for this problem.

✨

Key Insights

1Insight 1: Sorting the input array `nums` is crucial to efficiently identify and skip duplicate subsets during the backtracking process.
2Insight 2: The backtracking algorithm explores all possible combinations by recursively adding or not adding each element to a subset.
3Insight 3: To avoid duplicate subsets, a condition is added within the loop to skip over consecutive duplicate elements in the sorted array, ensuring each unique element is considered only once at each level of the recursion.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty input array `nums = []`. The solution correctly returns a list containing only an empty list `[[]]` which is the power set of an empty set.
Edge case 2: Input array with only duplicate elements `nums = [2, 2, 2]`. The solution correctly handles this by skipping duplicate elements at each level of recursion, generating subsets like `[]`, `[2]`, `[2, 2]`, and `[2, 2, 2]`.
Edge case 3: Input with negative numbers: `nums = [-1,0,1,2,-1,-4]`. The sorting step will rearrange and the duplicate skipping will work correctly on both negative and positive numbers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the input array before applying the backtracking algorithm. This will result in generating duplicate subsets.

✗Common mistake 2: Not correctly skipping duplicate elements within the backtracking loop. The condition `i > start && nums[i] == nums[i - 1]` is crucial for avoiding duplicate subsets. Forgetting the `i > start` part can cause elements after the first to be skipped when they are meant to be included at the top level recursion.

✗Common mistake 3: Modifying the original subset list instead of creating a copy when adding it to the results. Since lists are mutable objects, all subsets in the final result will end up being the same (the last one generated). This is especially important to note when coding in Java, C++ or Python.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Feature selection in machine learning. When building a model, we might want to consider all possible combinations of features. This problem can be modeled as finding all subsets of the set of available features.

🏦Real-world scenario 2: Combinatorial optimization. Many optimization problems involve finding the best combination of items to include in a solution. The subset generation problem is a fundamental building block for solving these types of problems. Example: Selecting items to put in a knapsack so that total weight doesn't exceed a limit and the total value is maximized.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Subsets II"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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