Introduction

In this blog post, we will explore how to solve the LeetCode problem "90" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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

Explanation

Here's a solution to the power set problem, optimized for efficiency and clarity, along with explanations:

• Sorting for Deduplication: Sort the input array nums. This crucial step enables us to efficiently skip duplicate elements during the subset generation process, ensuring that the final result doesn't contain duplicate subsets.

• Backtracking with Skipping: Employ a backtracking algorithm to explore all possible combinations. After including an element, recursively explore further combinations. The key optimization is to skip over consecutive duplicate elements to avoid generating redundant subsets.

• Iterative Construction: Iteratively build subsets by adding elements one by one. This ensures that we explore all possible combinations of elements from the input array.

• Time & Space Complexity: The solution has a time complexity of O(N 2^N) and a space complexity of O(N 2^N), where N is the number of elements in the input array.

Code

    def subsetsWithDup(nums):
    """
    Finds all possible subsets (power set) of an integer array that may contain duplicates,
    ensuring no duplicate subsets in the result.

    Args:
        nums: An integer array.

    Returns:
        A list of lists representing all unique subsets of nums.
    """

    nums.sort()  # Sort to handle duplicates effectively
    result = []
    subset = []

    def backtrack(index):
        result.append(subset[:])  # Add a copy of the current subset

        for i in range(index, len(nums)):
            # Skip duplicate numbers to avoid duplicate subsets
            if i > index and nums[i] == nums[i - 1]:
                continue

            subset.append(nums[i])
            backtrack(i + 1)
            subset.pop()  # Backtrack: remove the last element

    backtrack(0)
    return result