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Subsets II

Medium

Google

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Topics:

ArraysRecursion

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.

For example:

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

nums = [0] Output: [[],[0]]

Constraints:

1 <= nums.length <= 10
-10 <= nums[i] <= 10

Write a function to solve this problem efficiently. Consider edge cases and explain the time and space complexity of your solution.

Subsets II

Medium

Google

12 views

Topics:

ArraysRecursion

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.

For example:

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

nums = [0] Output: [[],[0]]

Constraints:

1 <= nums.length <= 10
-10 <= nums[i] <= 10

Write a function to solve this problem efficiently. Consider edge cases and explain the time and space complexity of your solution.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array? Are negative numbers, zero, or floating-point numbers possible?
Are there any constraints on the size of the input array? For example, is an empty array a valid input?
Does the order of the subsets in the output matter? And does the order of elements within each subset matter?
Are duplicate subsets allowed in the output, or should the output contain only unique subsets?
If the input array is already sorted, does that influence the approach I should take or any assumptions I can make?

Brute Force Solution

Approach

The brute force approach to finding all unique subsets involves exploring every possible combination. It's like meticulously trying on every outfit you could make from a set of clothes, even if some outfits look similar. We generate each potential subset and keep only the unique ones.

Here's how the algorithm would work step-by-step:

Consider the possibility of creating subsets without including any number from the input.
Then, generate subsets by including just the first number.
Next, create subsets including only the second number, and then the first and second numbers together.
Continue this process, making subsets by including only the third number, the first and third numbers, the second and third numbers, and then all three numbers together (if there are three numbers in the input).
Keep doing this, adding one number at a time and creating all possible combinations with the numbers you've seen so far.
After generating each subset, check if it is already in the list of subsets you've found. If it is, discard it. Otherwise, keep it.
Once you have considered all numbers and created every possible unique subset, you have your answer.

Code Implementation

def find_all_unique_subsets_brute_force(numbers):
    all_subsets = []

    # Iterate through all possible subsets
    for i in range(1 << len(numbers)):
        current_subset = []
        for j in range(len(numbers)):

            # Check if j-th bit is set in i
            if (i >> j) & 1:
                current_subset.append(numbers[j])

        # Sort the subset to handle duplicates.
        current_subset.sort()

        # Only add unique subsets
        if current_subset not in all_subsets:

            all_subsets.append(current_subset)

    return all_subsets

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm iterates through each number in the input array of size n, building subsets. In the worst-case scenario, it generates all possible subsets, which is 2^n. For each of these subsets, the algorithm performs a check to ensure it's unique, which involves comparing it against existing subsets. The unique check takes O(n) time in the worst case (comparing all elements of the subset), thus for each of the 2^n subsets it takes O(n) time to verify if it is unique. Therefore, the overall time complexity is O(n * 2^n).

Space Complexity

O(2^N * N) – The algorithm constructs all possible subsets of the input array of size N. In the worst case, there are 2^N possible subsets. Each subset can contain up to N elements. Therefore, we store, on average, the 2^N subsets, and each subset requires storage proportional to its length, which is at most N. Also, a list of subsets is maintained to check for duplicates, contributing to space usage by potentially storing 2^N subsets, each of size at most N. This leads to a space complexity of O(2^N * N).

Optimal Solution

Approach

The challenge is to find all possible groups of numbers from a list, including repetitions, but without creating duplicate groups in the final answer. The optimal solution leverages sorting and a strategic skipping technique to ensure each valid group is generated exactly once, avoiding redundancy.

Here's how the algorithm would work step-by-step:

First, sort the numbers. This allows us to easily group identical numbers together.
Start with an empty group. Then, consider adding each number to the group.
When deciding whether to add a number, check if it's the same as the previous number. If it is, and the previous number wasn't just added in the step before (meaning we're starting a new group with duplicate numbers), then skip it to avoid generating duplicate groups.
If the number is different from the previous number, or if the previous number was just added, then add it to the group, and continue making groups by adding other numbers. Also consider the option of not adding it and making groups with other remaining numbers
After exploring all possible groups formed by adding that number, remove it from the group (backtrack) and continue exploring other possibilities.
Repeat this process for each number in the original list, ensuring you don't create duplicate groups by skipping smartly.

Code Implementation

def find_all_unique_subsets(numbers):
    result = []
    subset = []
    numbers.sort()

    def backtrack(index):
        result.append(subset[:])

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

            subset.append(numbers[i])
            backtrack(i + 1)
            subset.pop()

    backtrack(0)
    return result

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible subsets of the input array of size n. At each element, it makes a decision: either include the element in the current subset or exclude it. This binary choice at each of the n elements leads to 2^n possible subsets. While the skipping technique avoids duplicates in the output, it doesn't fundamentally change the number of subsets explored in the worst case. Therefore, the time complexity is dominated by the exploration of the power set, resulting in O(2^n).

Space Complexity

O(N) – The auxiliary space is primarily determined by the depth of the recursion, which can go as deep as N, where N is the number of elements in the input list nums. Each level of recursion requires space to store the current subset being built. In the worst-case scenario, where no elements are pruned, the recursion depth is N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list immediately as there are no subsets to generate.
Input array with only one element	Return a list containing an empty list and a list containing the single element.
Input array with duplicate numbers and an empty subset	The algorithm should correctly handle duplicates to avoid generating identical subsets multiple times, typically using sorting and skipping duplicate numbers during recursion or iteration.
Input array with all identical numbers	The algorithm should still generate the correct subsets, including the empty set and sets of varying sizes with the same number.
Maximum size input array near memory limits	Ensure the algorithm doesn't use excessive memory, as the number of subsets grows exponentially with the input size, potentially leading to memory errors.
Negative numbers in the input array	The algorithm should handle negative numbers without issues, treating them the same as positive numbers during subset generation.
Integer overflow when calculating subset sizes if input numbers are large (though unlikely in this problem)	The subset sizes and values don't require large integer calculations that would overflow.
Large input array with many duplicates close to INT_MAX/INT_MIN	Sort the array to identify and skip duplicates, and handle the range of potential integer values correctly.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array? Are negative numbers, zero, or floating-point numbers possible?
Are there any constraints on the size of the input array? For example, is an empty array a valid input?
Does the order of the subsets in the output matter? And does the order of elements within each subset matter?
Are duplicate subsets allowed in the output, or should the output contain only unique subsets?
If the input array is already sorted, does that influence the approach I should take or any assumptions I can make?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider the possibility of creating subsets without including any number from the input.
Then, generate subsets by including just the first number.
Next, create subsets including only the second number, and then the first and second numbers together.
Continue this process, making subsets by including only the third number, the first and third numbers, the second and third numbers, and then all three numbers together (if there are three numbers in the input).
Keep doing this, adding one number at a time and creating all possible combinations with the numbers you've seen so far.
After generating each subset, check if it is already in the list of subsets you've found. If it is, discard it. Otherwise, keep it.
Once you have considered all numbers and created every possible unique subset, you have your answer.

Code Implementation

def find_all_unique_subsets_brute_force(numbers):
    all_subsets = []

    # Iterate through all possible subsets
    for i in range(1 << len(numbers)):
        current_subset = []
        for j in range(len(numbers)):

            # Check if j-th bit is set in i
            if (i >> j) & 1:
                current_subset.append(numbers[j])

        # Sort the subset to handle duplicates.
        current_subset.sort()

        # Only add unique subsets
        if current_subset not in all_subsets:

            all_subsets.append(current_subset)

    return all_subsets

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, sort the numbers. This allows us to easily group identical numbers together.
Start with an empty group. Then, consider adding each number to the group.
When deciding whether to add a number, check if it's the same as the previous number. If it is, and the previous number wasn't just added in the step before (meaning we're starting a new group with duplicate numbers), then skip it to avoid generating duplicate groups.
If the number is different from the previous number, or if the previous number was just added, then add it to the group, and continue making groups by adding other numbers. Also consider the option of not adding it and making groups with other remaining numbers
After exploring all possible groups formed by adding that number, remove it from the group (backtrack) and continue exploring other possibilities.
Repeat this process for each number in the original list, ensuring you don't create duplicate groups by skipping smartly.

Code Implementation

def find_all_unique_subsets(numbers):
    result = []
    subset = []
    numbers.sort()

    def backtrack(index):
        result.append(subset[:])

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

            subset.append(numbers[i])
            backtrack(i + 1)
            subset.pop()

    backtrack(0)
    return result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list immediately as there are no subsets to generate.
Input array with only one element	Return a list containing an empty list and a list containing the single element.
Input array with duplicate numbers and an empty subset	The algorithm should correctly handle duplicates to avoid generating identical subsets multiple times, typically using sorting and skipping duplicate numbers during recursion or iteration.
Input array with all identical numbers	The algorithm should still generate the correct subsets, including the empty set and sets of varying sizes with the same number.
Maximum size input array near memory limits	Ensure the algorithm doesn't use excessive memory, as the number of subsets grows exponentially with the input size, potentially leading to memory errors.
Negative numbers in the input array	The algorithm should handle negative numbers without issues, treating them the same as positive numbers during subset generation.
Integer overflow when calculating subset sizes if input numbers are large (though unlikely in this problem)	The subset sizes and values don't require large integer calculations that would overflow.
Large input array with many duplicates close to INT_MAX/INT_MIN	Sort the array to identify and skip duplicates, and handle the range of potential integer values correctly.