Subsets Leetcode Problem 78 [Python Solution]

In this post, we’ll tackle the Subsets problem from LeetCode, which is categorized under the “Backtracking” section.

This is a medium difficulty problem and a common interview question, especially at tech giants like Facebook.

Problem Overview

The problem statement goes as follows: Given an integer array nums containing unique elements, you are tasked with returning all possible subsets (the power set).

The key point here is that the solution set must not contain duplicate subsets, and you can return the solutions in any order.

Example:

Let’s illustrate this with an example:

• Input: nums = [1,2,3]
• Output: [[], [1], [2], [1,2], [3], [1,3], [2,3], [1,2,3]]

In the example, we have an array [1, 2, 3], and we need to generate all possible subsets.

The result contains all unique subsets, and the order does not matter.

Understanding the Constraints

Before delving into the solution, let’s understand the constraints:

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10
• All the numbers in nums are unique.

Given these constraints, the brute force approach is acceptable, but it’s essential to keep in mind that the number of subsets grows exponentially with the size of the input, making efficiency a crucial concern.

Subsets LeetCode Problem Solution

Bruteforce Approach

The most straightforward way to solve this problem is to use a backtracking approach.

For each element in the nums array, we have two choices: either include it in the current subset or skip it.

We’ll build all possible subsets by exploring both choices recursively.

Here’s a Python function that implements this approach:

def subsets(nums):
    res = []  # The list to store subsets

    def backtrack(start=0, subset=[]):
        res.append(subset[:])  # Add a copy of the current subset to the result
        for i in range(start, len(nums)):
            subset.append(nums[i])  # Include the current element
            backtrack(i + 1, subset)  # Recursively generate subsets
            subset.pop()  # Backtrack by removing the current element

    backtrack()  # Start the backtracking process
    return res

The backtrack function takes two parameters: start (the starting index for the current subset) and subset (the current subset being built).

We start from the beginning of the nums array and consider two choices for each element: either include it or skip it.

We use backtracking to explore both options.

Efficient Approach with Backtracking

The reasoning behind this approach is that for each element in the array, we have two choices: to include it in the current subset or skip it.

By exploring both options recursively, we can generate all possible subsets without duplicates.

Python Code Solution

Here is the efficient Python code solution for the Subsets problem:

def subsets(nums):
    res = []  # The list to store subsets

    def backtrack(start=0, subset=[]):
        res.append(subset[:])  # Add a copy of the current subset to the result
        for i in range(start, len(nums)):
            subset.append(nums[i])  # Include the current element
            backtrack(i + 1, subset)  # Recursively generate subsets
            subset.pop()  # Backtrack by removing the current element

    backtrack()  # Start the backtracking process
    return res

This solution efficiently generates all unique subsets of the given nums array.

Time and Space Complexity

Now, let’s analyze the time and space complexity of our solution.

Time Complexity:

The time complexity of our solution is O(2^n), where n is the length of the nums array.

This is because, for each element in the array, we have two choices (include it or skip it).

Since there are 2^n subsets in total, we must explore all of them.

Space Complexity:

The space complexity of our solution is O(n), where n is the length of the nums array.

This is due to the space required for the subset list, which can have a maximum of n elements in it.

Reasoning Behind Our Approach

The reasoning behind our approach lies in understanding that generating all possible subsets is inherently an exponential problem.

The number of subsets grows with the size of the input, and we cannot avoid this exponential growth.

Therefore, backtracking, which explores all possible choices, is a suitable approach for this problem.

By using backtracking, we efficiently explore the decision tree of including or excluding each element, and we ensure that we don’t include duplicate subsets in our result.

Related Interview Questions By Company:

• Longest Increasing Subsequence LeetCode
• Check If A String Contains All Binary Codes Of Size K LeetCode
• Design Twitter LeetCode

Related Interview Questions By Difficulty:

• Combination Sum II LeetCode
• Matchsticks To Square LeetCode
• Letter Combinations Of A Phone Number LeetCode

Conclusion

In this blog post, we’ve tackled the Subsets problem from LeetCode, providing a Python solution.

We’ve discussed the problem’s constraints, explained the backtracking approach, and presented the efficient Python code solution.

We’ve also analyzed the time and space complexity of our solution.

If you found this post helpful or have any questions, feel free to comment, ask questions, make suggestions, or share the content.

Learning and discussing algorithms and data structures is a collaborative journey, and your input is highly valued.

To try the problem yourself, visit the LeetCode Subsets Problem.

Happy coding!