Naive Solution: Brute Force (with Duplicates)

The most straightforward approach is to generate all possible subsets without considering duplicates initially and then remove the duplicate subsets at the end.

1. Generate all subsets: Iterate through all possible combinations of elements from the input array nums. This can be achieved using nested loops or recursion.
2. Store subsets: Store each generated subset in a list or set.
3. Remove duplicates: After generating all subsets, iterate through the list and remove any duplicate subsets. This typically involves converting each subset to a comparable form (e.g., a tuple) and using a set to track unique subsets.

Big O Analysis

• Time Complexity: O(N * 2^N) to generate all subsets, where N is the number of elements in nums. Additionally, removing duplicates can take O(M²) where M = 2^N in the worst case, leading to a combined complexity of O(N * 2^N + M²). The actual complexity of removing duplicates depends on the data structure used and the method applied. In the worst case, M == 2^N, so O(2^2N) for duplicate removals.
• Space Complexity: O(2^N) to store all possible subsets. Additionally, if extra space is used for duplicate removals, this can increase the space complexity further.

Edge Cases

• Empty input array: Should return a list containing only an empty list.
• Input array with only one element: Should return a list containing an empty list and a list containing the single element.
• Input array with all duplicate elements: Ensure the code handles this efficiently without generating redundant computations.

Example Code (Python)

def subsets_with_duplicates_brute_force(nums):
    subsets = []
    for i in range(1 << len(nums)):
        subset = []
        for j in range(len(nums)):
            if (i >> j) & 1:
                subset.append(nums[j])
        subsets.append(subset)

    unique_subsets = []
    for subset in subsets:
        subset.sort()
        if subset not in unique_subsets:
            unique_subsets.append(subset)

    return unique_subsets

# Example Usage
nums = [1, 2, 2]
result = subsets_with_duplicates_brute_force(nums)
print(result)

Optimal Solution: Backtracking with Skipping Duplicates

A more efficient approach is to use backtracking. The key optimization here is to avoid generating duplicate subsets during the backtracking process itself. This is typically done by sorting the input array and skipping over duplicate elements during the recursive calls.

1. Sort the input array: Sorting allows easy identification of duplicate elements.
2. Backtracking: Recursively build subsets. At each step:
   • Include the current element in the subset and make a recursive call.
   • Exclude the current element. If the current element is the same as the previous element (and the previous element was not included in the subset), skip it to avoid generating duplicate subsets.

Big O Analysis

• Time Complexity: O(N * 2^N) - While still exponential due to the nature of generating all subsets, the algorithm avoids redundant computations, making it more efficient than the brute-force approach with post-processing for duplicate removal. The 'N' factor comes from potentially copying the current subset in each recursive call.
• Space Complexity: O(N) - Primarily due to the recursion stack. The depth of the recursion can be at most N.

Edge Cases

• Empty input array: Should return a list containing only an empty list.
• Input array with only one element: Should return a list containing an empty list and a list containing the single element.
• Input array with all duplicate elements: The skipping mechanism should handle this case efficiently.

Example Code (Python)

def subsets_with_duplicates_optimal(nums):
    result = []
    nums.sort()

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

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

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

    backtrack(0, [])
    return result

# Example Usage
nums = [1, 2, 2]
result = subsets_with_duplicates_optimal(nums)
print(result)