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Subsets

Medium

Bloomberg LP

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Topics:

ArraysRecursionBit Manipulation

Given an integer array nums of unique elements, 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 = [1,2,3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Example 2:

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

Constraints:

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

Subsets

Medium

Bloomberg LP

1 view

Topics:

ArraysRecursionBit Manipulation

Given an integer array nums of unique elements, 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 = [1,2,3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Example 2:

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

Constraints:

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

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 maximum size of the input array `nums`?
Can the input array `nums` contain negative numbers?
Is the order of subsets in the output important, or can they be in any order?
Is the order of elements within each subset important, or can they be in any order?
Is the input array `nums` guaranteed to be non-null?

Brute Force Solution

Approach

The goal is to find all possible combinations of items from a given set. A brute force approach means we're going to explore every single possible combination. Think of it as exhaustively trying everything.

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

Start with an empty group. This is always a possibility: choosing nothing.
Then, consider each item individually. Each item, by itself, is a possible group.
Next, consider every pair of items. Take each pair and see if it forms a valid group.
After pairs, look at every possible group of three items. Then groups of four, and so on.
Keep going until you've considered a group that includes all the items. This is also a valid group.
By the end, you will have tried every possible combination of items, giving you all the subsets.

Code Implementation

def find_all_subsets_brute_force(input_set):
    all_subsets = []

    number_of_elements = len(input_set)

    # Start with an empty set, which is always a subset
    all_subsets.append([])

    for subset_size in range(1, number_of_elements + 1):
        for i in range(1 << number_of_elements):
            # Generate all possible combinations of given size
            if bin(i).count('1') == subset_size:
                current_subset = []

                for j in range(number_of_elements):
                    # Check if the j-th bit is set in the current combination
                    if (i >> j) & 1:
                        current_subset.append(input_set[j])

                all_subsets.append(current_subset)

    return all_subsets

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm systematically generates subsets of all possible sizes from the empty set to the set containing all n elements. Generating subsets of size k from a set of n elements takes n choose k operations. Summing this for all possible subset sizes (from k=0 to k=n) results in the sum of binomial coefficients, which is equal to 2^n. Therefore, the time complexity is O(2^n) because we are essentially exploring every possible combination of elements to form subsets.

Space Complexity

O(N) – The brute force approach described involves constructing all possible subsets. Each subset is stored as a separate group. In the worst case, where N is the number of items in the input set, we could potentially store a group containing all N items. While the algorithm generates 2^N subsets, only one subset of size N may be stored at any given time, such as when considering the subset with all items. Therefore, the space complexity is determined by the largest subset being constructed, which can contain up to N items, resulting in O(N) auxiliary space.

Optimal Solution

Approach

The best way to find all subsets is to think about making a decision for each item: do we include it in a subset, or not? We build up all possible subsets by systematically going through each item and considering both possibilities.

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

Start with an empty subset.
Consider the first item. We now have two choices: either include it in our current subsets, or don't.
If we include it, we add a copy of all the current subsets with the item added to each one.
If we don't include it, we keep the current subsets as they are.
Now we have a new collection of subsets, representing all choices made so far.
Repeat this process for each remaining item in the original set.
At each step, we're doubling the number of subsets as we decide whether or not to include the current item.
Once we've gone through all the items, the final collection of subsets represents all possible combinations, including the empty set and the set itself.

Code Implementation

def find_all_subsets(input_set):
    all_subsets = [[]] 

    for current_number in input_set:
        # Iterate through each number in the input set

        new_subsets = []

        for existing_subset in all_subsets:
            # Create new subsets including current number
            new_subsets.append(existing_subset + [current_number])

        # Add the new subsets to the collection
        all_subsets.extend(new_subsets)

    return all_subsets

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm iterates through each of the n elements in the input set. For each element, it effectively doubles the number of subsets accumulated so far. Since there are 2^n possible subsets in total, constructing each subset takes O(n) time (to copy the existing subset and potentially add the current element). Therefore, the overall time complexity is O(n * 2^n).

Space Complexity

O(2^N) – The algorithm generates all possible subsets of the input set. At each step, we are doubling the number of subsets we store. This means we are creating a new collection of subsets, where each subset potentially has a size of up to N (the number of elements in the input set). Therefore, the space required to store all 2^N subsets is proportional to N * 2^N in the worst case since each subset can have at most N elements. However, when considering space complexity, we usually focus on the number of subsets and ignore the size of each subset, so the space complexity is O(2^N).

Edge Cases

Case	How to Handle
Null or empty input array	Return a list containing only an empty list to represent the empty subset.
Input array with a single element	Return a list containing an empty list and a list containing the single element.
Large input array size (performance considerations)	Ensure the algorithm's time complexity is acceptable, given that the number of subsets grows exponentially (O(2^n)).
Array containing negative numbers	The standard subset generation algorithms should handle negative numbers without any special modifications.
Array containing zero	The standard subset generation algorithms should handle zero without any special modifications.
Integer overflow (if using bit manipulation)	Consider the maximum size of the input array and whether the bit representation of subsets will exceed the integer limits and switch to a different subset generation algorithm if needed.
Inputs close to maximum integer values	Standard subset generation will handle extreme boundary values as they are, as long as we're not performing arithmetic operations prone to overflows.
All numbers are identical (though the problem statement specifies 'unique integers')	Despite being unlikely due to constraints, standard subset generation will work since we only care about if an element is present or not and not about distinct elements.

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 maximum size of the input array `nums`?
Can the input array `nums` contain negative numbers?
Is the order of subsets in the output important, or can they be in any order?
Is the order of elements within each subset important, or can they be in any order?
Is the input array `nums` guaranteed to be non-null?

Brute Force Solution

Approach

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

Start with an empty group. This is always a possibility: choosing nothing.
Then, consider each item individually. Each item, by itself, is a possible group.
Next, consider every pair of items. Take each pair and see if it forms a valid group.
After pairs, look at every possible group of three items. Then groups of four, and so on.
Keep going until you've considered a group that includes all the items. This is also a valid group.
By the end, you will have tried every possible combination of items, giving you all the subsets.

Code Implementation

def find_all_subsets_brute_force(input_set):
    all_subsets = []

    number_of_elements = len(input_set)

    # Start with an empty set, which is always a subset
    all_subsets.append([])

    for subset_size in range(1, number_of_elements + 1):
        for i in range(1 << number_of_elements):
            # Generate all possible combinations of given size
            if bin(i).count('1') == subset_size:
                current_subset = []

                for j in range(number_of_elements):
                    # Check if the j-th bit is set in the current combination
                    if (i >> j) & 1:
                        current_subset.append(input_set[j])

                all_subsets.append(current_subset)

    return all_subsets

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start with an empty subset.
Consider the first item. We now have two choices: either include it in our current subsets, or don't.
If we include it, we add a copy of all the current subsets with the item added to each one.
If we don't include it, we keep the current subsets as they are.
Now we have a new collection of subsets, representing all choices made so far.
Repeat this process for each remaining item in the original set.
At each step, we're doubling the number of subsets as we decide whether or not to include the current item.
Once we've gone through all the items, the final collection of subsets represents all possible combinations, including the empty set and the set itself.

Code Implementation

def find_all_subsets(input_set):
    all_subsets = [[]] 

    for current_number in input_set:
        # Iterate through each number in the input set

        new_subsets = []

        for existing_subset in all_subsets:
            # Create new subsets including current number
            new_subsets.append(existing_subset + [current_number])

        # Add the new subsets to the collection
        all_subsets.extend(new_subsets)

    return all_subsets

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return a list containing only an empty list to represent the empty subset.
Input array with a single element	Return a list containing an empty list and a list containing the single element.
Large input array size (performance considerations)	Ensure the algorithm's time complexity is acceptable, given that the number of subsets grows exponentially (O(2^n)).
Array containing negative numbers	The standard subset generation algorithms should handle negative numbers without any special modifications.
Array containing zero	The standard subset generation algorithms should handle zero without any special modifications.
Integer overflow (if using bit manipulation)	Consider the maximum size of the input array and whether the bit representation of subsets will exceed the integer limits and switch to a different subset generation algorithm if needed.
Inputs close to maximum integer values	Standard subset generation will handle extreme boundary values as they are, as long as we're not performing arithmetic operations prone to overflows.
All numbers are identical (though the problem statement specifies 'unique integers')	Despite being unlikely due to constraints, standard subset generation will work since we only care about if an element is present or not and not about distinct elements.