What is the time complexity of Subsets?

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

What data structures are used to solve Subsets?

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

What are the key insights for solving Subsets?

Insight 1: Backtracking is a suitable technique because we explore all possible combinations by making choices (include an element or not) and then undoing those choices to explore other paths. Insight 2: Each element in 'nums' has two possibilities: either be included in the current subset or not included. This leads to 2^n possible subsets.

Which companies ask LeetCode 78. Subsets in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, ByteDance and 15 more companies.

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Subsets - LeetCode 78 Solution

Q: How to solve Subsets?

The provided solution uses a backtracking algorithm to generate all subsets. Backtracking involves exploring all possible choices recursively. For each element in the input array, we either include it in the current subset or exclude it. The 'backtrack' function maintains a 'path' (current subset) and explores possible subsets by adding elements from the input array 'nums' to the 'path'. After exploring a path, the function removes the last added element (backtracks) to explore other possibilities. The 'start' parameter ensures that we only consider elements from the current index onwards, preventing duplicate subsets.

Subsets - Complete Solution Guide

Subsets is LeetCode problem 78, 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 of unique elements, 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 = [1,2,3] Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]] Example 2: Input: nums = [0] Output: [[],[0]] Constraints: 1 <= nums.length <= 10 -10 <= nums[i] <= 10 All the numbers of nums are unique.

Detailed Explanation

The problem asks us to generate all possible subsets (also known as the power set) of a given integer array 'nums' containing unique elements. A subset is a selection of elements from the original array, and it can include any number of elements from zero (the empty set) to all elements (the original set itself). The output should be a list of lists, where each inner list represents a unique subset of 'nums'. The order of the subsets in the output doesn't matter.

Solution Approach

The provided solution uses a backtracking algorithm to generate all subsets. Backtracking involves exploring all possible choices recursively. For each element in the input array, we either include it in the current subset or exclude it. The 'backtrack' function maintains a 'path' (current subset) and explores possible subsets by adding elements from the input array 'nums' to the 'path'. After exploring a path, the function removes the last added element (backtracks) to explore other possibilities. The 'start' parameter ensures that we only consider elements from the current index onwards, preventing duplicate subsets.

Step-by-Step Algorithm

Step 1: Initialize an empty list 'res' to store all the generated subsets.
Step 2: Define a recursive 'backtrack' function that takes 'start' index, the input 'nums', current 'path', and the result 'res' list as arguments.
Step 3: Inside 'backtrack', add a copy of the current 'path' to the 'res' list. This represents a valid subset.
Step 4: Iterate through the 'nums' array from the 'start' index to the end.
Step 5: For each element at index 'i', add it to the current 'path'.
Step 6: Recursively call 'backtrack' with 'i + 1' as the new 'start' index. This explores the possibility of including the current element in the subset.
Step 7: After the recursive call returns, remove the last added element from the 'path' (backtracking). This explores the possibility of excluding the current element from the subset.
Step 8: The base case for the recursion is when all elements in 'nums' have been considered. The empty path is added initially, and after exploring all choices, the 'res' list contains all possible subsets.
Step 9: Return the 'res' list containing all generated subsets.

Key Insights

Insight 1: Backtracking is a suitable technique because we explore all possible combinations by making choices (include an element or not) and then undoing those choices to explore other paths.
Insight 2: Each element in 'nums' has two possibilities: either be included in the current subset or not included. This leads to 2^n possible subsets.
Insight 3: To avoid generating duplicate subsets, we use a 'start' index in the backtracking function. This ensures that we only consider elements from the current index onwards, preventing redundant combinations.

Complexity Analysis

Time Complexity: O(n * 2^n)

Space Complexity: O(n)

Topics

This problem involves: Backtracking, Bit Manipulation, Array.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, ByteDance, Coupang, Fiverr, IBM, Infosys, Mastercard, Meta, Microsoft, Salesforce, Samsung, TikTok, Uber, Wix, Yahoo, Zoho, tcs.

Subsets | LeetCode 78

Medium

Backtracking Bit Manipulation Array

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        backtrack(0, nums, new ArrayList<>(), res);
        return res;
    }

    private void backtrack(int start, int[] nums, List<Integer> path, List<List<Integer>> res) {
        res.add(new ArrayList<>(path));
        for (int i = start; i < nums.length; i++) {
            path.add(nums[i]);
            backtrack(i + 1, nums, path, res);
            path.remove(path.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 78: Subsets - Detailed explanation and solution approaches.

Topics

Backtracking Bit Manipulation Array

Companies

AdobeAmazonAppleBloombergByteDanceCoupangFiverrIBM

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:78

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: Initialize an empty list 'res' to store all the generated subsets.
2
Step 2: Define a recursive 'backtrack' function that takes 'start' index, the input 'nums', current 'path', and the result 'res' list as arguments.
3
Step 3: Inside 'backtrack', add a copy of the current 'path' to the 'res' list. This represents a valid subset.
4
Step 4: Iterate through the 'nums' array from the 'start' index to the end.
5
Step 5: For each element at index 'i', add it to the current 'path'.
6
Step 6: Recursively call 'backtrack' with 'i + 1' as the new 'start' index. This explores the possibility of including the current element in the subset.
7
Step 7: After the recursive call returns, remove the last added element from the 'path' (backtracking). This explores the possibility of excluding the current element from the subset.
8
Step 8: The base case for the recursion is when all elements in 'nums' have been considered. The empty path is added initially, and after exploring all choices, the 'res' list contains 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)

📦 Linear space - Memory grows with input size.

📝

Why?

Time Complexity: There are 2^n possible subsets. For each subset, we need to copy the elements into a new list, which takes O(n) time. Therefore, the overall time complexity is O(n * 2^n). The 'n' factor comes from copying the subset into the result list. Space Complexity: The maximum depth of the recursion is 'n' (the number of elements in 'nums'). This corresponds to the space used by the call stack during backtracking. Additionally, the 'path' variable also uses O(n) space in the worst case where all elements are included in the current subset. Therefore, the space complexity is O(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

Bit Manipulation

We can represent each subset using a bitmask. If the i-th bit is set, the i-th element of the input array is included in the subset. We iterate through all possible bitmasks from 0 to 2^n - 1, creating a subset for each bitmask.

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

Trade-offs: Bit manipulation can be slightly faster in practice due to lower overhead compared to recursion. However, it might be less intuitive for beginners than backtracking. The space complexity can vary implementation to implementation if extra space is used to store the intermediate results. Both approaches yield the same subsets, so code clarity might be a more important deciding factor.

✨

Key Insights

1Insight 1: Backtracking is a suitable technique because we explore all possible combinations by making choices (include an element or not) and then undoing those choices to explore other paths.
2Insight 2: Each element in 'nums' has two possibilities: either be included in the current subset or not included. This leads to 2^n possible subsets.
3Insight 3: To avoid generating duplicate subsets, we use a 'start' index in the backtracking function. This ensures that we only consider elements from the current index onwards, preventing redundant combinations.

📚

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: If the input 'nums' is empty, the solution will return a list containing only the empty set [[]]. This is the correct behavior.
Edge case 2: Input array with one element: If the input 'nums' contains only one element, the solution will return a list containing the empty set [] and the set containing the single element [[], [element]]. This is also the expected behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not copying the 'path' before adding it to the 'res' list. If you add 'path' directly, all subsets in 'res' will point to the same list, and the final result will be incorrect as modifications to 'path' affect all stored subsets. Using `.copy()` in Python or `new ArrayList<>(path)` in Java or `vector<int> path` in C++ creates a copy to avoid this.

✗Common mistake 2: Not backtracking correctly by removing the last added element from the 'path'. This will lead to incorrect subsets being generated.

✗Common mistake 3: Forgetting the base case. The base case implicitly happens by the end of the iteration in the backtrack function where start >= nums.length, and where we add the path which is an empty subset. The start is incremented at each recursive call

✅

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 machine learning model, one might want to try different combinations of features to find the optimal set that maximizes model performance. Generating all subsets of features can be useful in this scenario.

🏦Real-world scenario 2: Generating combinations for product recommendations: E-commerce platforms can use subsets to suggest product bundles or complementary items to customers based on their past purchases.

💡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

Backtracking = DFS + pruning. The key is identifying when to backtrack. Always restore state after recursion.

🎯 Interview Tip

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

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 18, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

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"?

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

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