Subsets II Leetcode Problem 90 [Python Solution]

Today, we’re going to code out the Subsets II problem on LeetCode.

This is a follow-up to the original “Subsets” problem, and it comes with a little twist.

We’re going to dive into the details, explore the problem, develop a solution, and provide a Python solution.

If you’re a beginner, don’t worry, we’ll walk you through every step.

So, let’s get started!

Problem Overview

The Subsets II problem on LeetCode goes like this: Given an integer array nums that may contain duplicates, our task is to return all possible subsets, also known as the power set.

However, there’s a catch.

The solution set must not contain duplicate subsets, and we can return the solution in any order.

Let’s take a look at an example:

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

Example 2:
Input: nums = [0]
Output: [[],[0]]

Understanding the Constraints

Before we jump into the solution, let’s understand the constraints of this problem:

• The length of the nums array is between 1 and 10.
• The values in the nums array can range from -10 to 10. Now that we have a clear picture of the problem, let’s dive into the solution.

Subsets II LeetCode Problem Solution

To solve this problem, we’ll use a backtracking approach.

Backtracking is a technique for finding all possible solutions by incrementally building a solution and undoing it if we determine it’s not valid.

1. Bruteforce Approach

The brute force approach for this problem is to consider all possible subsets and then filter out the duplicates.

Here’s a Python code snippet to illustrate this approach:

def subsetsWithDup(nums):
    res = []
    nums.sort()

    def backtrack(i, subset):
        if i == len(nums):
            res.append(subset[:])  # Make a copy of the subset
            return

        # Include the current element
        subset.append(nums[i])
        backtrack(i + 1, subset)
        subset.pop()

        # Skip the duplicates
        while i + 1 < len(nums) and nums[i] == nums[i + 1]:
            i += 1

        # Exclude the current element
        backtrack(i + 1, subset)

    backtrack(0, [])
    return res

In this solution, we maintain a subset to build our subsets incrementally.

We sort the input array nums to make it easier to handle duplicates.

We use the backtrack function to explore the two choices: include the current element or skip it while handling duplicates to avoid duplicate subsets.

The backtrack function recursively explores all possible subsets and appends them to the res array.

The key insight here is that when we encounter duplicates, we skip them in one branch of the recursion to avoid duplicates in the result.

This approach is intuitive and correct, but it may not be the most efficient one.

Let’s analyze the time and space complexity.

Time and Space Complexity

The time complexity of this solution is approximately O(N * 2^N), where N is the number of elements in the nums array.

The reason is that we have two choices for each element (include or exclude), and we go through all possible subsets.

Sorting the input array takes O(N * log(N)), which is insignificant compared to the exponential part.

The space complexity is O(N) for the recursion stack and O(N * 2^N) for the result list to store all subsets.

Therefore, the overall space complexity is O(N * 2^N).

Reasoning Behind Our Approach

We chose a backtracking approach to solve the Subsets II problem because it allows us to systematically explore all possible subsets while avoiding duplicates.

Sorting the input array simplifies the process of handling duplicates, and our backtracking strategy efficiently generates the result.

By choosing this approach, we strike a balance between correctness and efficiency.

Related Interview Questions By Company:

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Related Interview Questions By Difficulty:

• Partition To K Equal Sum Subsets LeetCode
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Conclusion

In this blog post, we delved into the Subsets II problem on LeetCode.

We discussed the problem statement, examined the constraints, and developed a Python solution using a backtracking approach.

Our solution allows us to generate all possible subsets while avoiding duplicates.

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

We hope this explanation and solution were helpful to you, especially if you’re a beginner.

If you have any questions, suggestions, or want to discuss the problem further, please feel free to leave a comment.

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LeetCode Problem Link

Remember, understanding these problems takes practice, so keep working on different examples to strengthen your problem-solving skills.

Happy coding!