Jobs

Interviews

Discussions

The Number of Beautiful Subsets

Medium

Asked by:

24 views

Topics:

ArraysRecursionBit Manipulation

You are given an array nums of positive integers and a positive integer k.

A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k.

Return the number of non-empty beautiful subsets of the array nums.

A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [2,4,6], k = 2
Output: 4
Explanation: The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].

Example 2:

Input: nums = [1], k = 1
Output: 1
Explanation: The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].

Constraints:

1 <= nums.length <= 18
1 <= nums[i], k <= 1000

Try coding it in LeetCode

The Number of Beautiful Subsets

Medium

Asked by:

24 views

Topics:

ArraysRecursionBit Manipulation

You are given an array nums of positive integers and a positive integer k.

A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k.

Return the number of non-empty beautiful subsets of the array nums.

A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

Example 1:

Input: nums = [2,4,6], k = 2
Output: 4
Explanation: The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].

Example 2:

Input: nums = [1], k = 1
Output: 1
Explanation: The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].

Constraints:

1 <= nums.length <= 18
1 <= nums[i], k <= 1000

Try coding it in LeetCode

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 for the numbers in the input array?
Can the input array contain negative numbers or zero?
What defines a 'beautiful subset' exactly? Can you provide an example beyond the problem description to ensure I understand the condition?
Are there any constraints on the size of the input array, and should I be concerned about potential integer overflow when calculating the number of subsets?
If the input array is empty, what should the function return?

Brute Force Solution

Approach

The brute force strategy is all about exploring every single possibility to find the answer. We'll consider every possible subset of the numbers given to us. For each subset, we will check if it meets the criteria to be a 'beautiful' subset.

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

First, let's consider the possibility of having an empty set (no numbers at all). We'll keep track of this as one possible subset.
Next, let's consider each individual number in our list. Each one of them forms a subset by itself.
Now, let's consider all possible pairs of numbers from our list.
After that, we'll consider all possible groups of three numbers from the list.
We continue this process, making sure to try every single combination of numbers, all the way up to using all the numbers in the list together in one big set.
For each of these subsets, we'll carefully check if it is a 'beautiful' subset. This means we'll look at all the pairs of numbers within the subset and confirm that their difference isn't a forbidden number.
Finally, we count all the subsets that meet our 'beautiful' criteria. That number is our answer.

Code Implementation

def the_number_of_beautiful_subsets(numbers, difference):
    number_of_beautiful_subsets = 0
    
    # Iterate through all possible subsets
    for i in range(1 << len(numbers)):
        subset = []
        for j in range(len(numbers)):
            if (i >> j) & 1:
                subset.append(numbers[j])

        # Check if the subset is beautiful
        is_beautiful = True
        #An empty set is defined as beautiful
        if not subset:
            number_of_beautiful_subsets += 1
            continue
        
        for first_index_subset in range(len(subset)):
            for second_index_subset in range(first_index_subset + 1, len(subset)):
                #Check the differences between each pair in the subset
                if abs(subset[first_index_subset] - subset[second_index_subset]) == difference:
                    is_beautiful = False
                    break
            if not is_beautiful:
                break

        # Increment the count if beautiful
        if is_beautiful:
            number_of_beautiful_subsets += 1

    return number_of_beautiful_subsets

Big(O) Analysis

Time Complexity

O(2^n * n^2) – The brute force approach generates all possible subsets of the input array. An array of size n has 2^n subsets. For each subset, we need to check if it is beautiful. In the worst case, for a subset of size n, we need to compare each pair of numbers to ensure the absolute difference is not k. Checking each pair takes O(n^2) time. Therefore, the overall time complexity is O(2^n * n^2) where we generate 2^n subsets and for each we check at most n^2 pairs.

Space Complexity

O(1) – The provided brute force approach primarily iterates through subsets and checks a condition. It doesn't appear to use any auxiliary data structures that scale with the input size N (the number of elements in the input list). It is implied that the differences within the subsets are checked using a constant amount of extra memory. Thus, the space complexity is constant.

Optimal Solution

Approach

The best way to solve this problem is to realize we can build up the answer incrementally. We'll look at each number one by one and decide whether to include it in a beautiful subset, based on whether it conflicts with any numbers already chosen.

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

First, create a way to remember which numbers are already part of a beautiful subset.
Go through the list of numbers one at a time.
For each number, check if adding it to the current subset would violate the 'beautiful' rule (meaning it shouldn't have a difference of exactly 1 with any number already chosen).
If adding the number is okay, we have two choices: either include it in the current subset, or skip it and leave it out.
Try both possibilities and count how many beautiful subsets can be created in each case.
Combine the counts from both possibilities (including the number and excluding the number) to get the total number of beautiful subsets that can be formed from this point forward.
Use the final count to determine the total number of beautiful subsets for the entire original list of numbers.

Code Implementation

def the_number_of_beautiful_subsets(numbers):
    seen_numbers = set()

    def count_beautiful_subsets(index):
        # Base case: If we've processed all numbers, return 1 (empty set).
        if index == len(numbers):
            return 1

        current_number = numbers[index]
        # Check if adding the number violates the beautiful subset property.
        is_beautiful = True
        if (current_number - 1) in seen_numbers or (current_number + 1) in seen_numbers:
            is_beautiful = False

        count = 0
        # Option 1: Exclude the current number.
        count += count_beautiful_subsets(index + 1)

        # Option 2: Include the current number if it's allowed.
        if is_beautiful:
            seen_numbers.add(current_number)
            count += count_beautiful_subsets(index + 1)

            # Backtrack: Remove the number to explore other possibilities.
            seen_numbers.remove(current_number)

        return count

    # Subtract 1 to exclude the empty subset.
    return count_beautiful_subsets(0) - 1

Big(O) Analysis

Time Complexity

O(2^n) – The described solution involves a recursive approach where, for each of the n numbers in the input, we explore two possibilities: include it in a beautiful subset or exclude it. This branching creates a binary tree of decisions. In the worst-case scenario, we explore all possible combinations of including and excluding elements, leading to 2^n possible subsets. Therefore, the time complexity is O(2^n).

Space Complexity

O(N) – The space complexity stems primarily from the 'remember which numbers are already part of a beautiful subset' requirement. This would likely be implemented using a set or a boolean array of size N, where N is the number of elements in the input list, to track the numbers included in the current subset. Additionally, recursive calls will contribute to the call stack space. In the worst-case scenario, without memoization or iterative approach, the recursion depth could reach N, leading to O(N) space for the call stack. Combining these factors, the dominant space usage is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 1 as an empty set is always a beautiful subset.
Array containing only the number 0	Treat 0 as any other number; if it is the only number, return 2 (empty set and {0})
Array with a single element	Return 2 (empty set and the set containing the single element).
Array with all identical elements	Iterate to check each number's presence and handle the identical number efficiently.
Array containing a wide range of number values	The solution should not overflow when calculating sums of elements, potentially needing 64-bit integers.
Input array with a very large number of elements	Optimize the solution for time complexity, potentially using dynamic programming or memoization to avoid redundant calculations to prevent exceeding time constraints.
Array contains negative numbers or very large positive numbers	Handle negative numbers by not allowing them in the selection or adjust the difference check accordingly, and large numbers to prevent integer overflow during the 'different by one' check.
Input array contains duplicate values	The solution should correctly account for duplicate values when forming subsets, counting each unique set only once, which can be done with the standard iterative subset enumeration logic.

Empty input array

How to Handle:

Return 1 as an empty set is always a beautiful subset.

Array containing only the number 0

How to Handle:

Treat 0 as any other number; if it is the only number, return 2 (empty set and {0})

Array with a single element

How to Handle:

Return 2 (empty set and the set containing the single element).

Array with all identical elements

How to Handle:

Iterate to check each number's presence and handle the identical number efficiently.

Array containing a wide range of number values

How to Handle:

The solution should not overflow when calculating sums of elements, potentially needing 64-bit integers.

Input array with a very large number of elements

How to Handle:

Optimize the solution for time complexity, potentially using dynamic programming or memoization to avoid redundant calculations to prevent exceeding time constraints.

Array contains negative numbers or very large positive numbers

How to Handle:

Handle negative numbers by not allowing them in the selection or adjust the difference check accordingly, and large numbers to prevent integer overflow during the 'different by one' check.

Input array contains duplicate values

How to Handle:

The solution should correctly account for duplicate values when forming subsets, counting each unique set only once, which can be done with the standard iterative subset enumeration logic.

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 for the numbers in the input array?
Can the input array contain negative numbers or zero?
What defines a 'beautiful subset' exactly? Can you provide an example beyond the problem description to ensure I understand the condition?
Are there any constraints on the size of the input array, and should I be concerned about potential integer overflow when calculating the number of subsets?
If the input array is empty, what should the function return?

Brute Force Solution

Approach

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

First, let's consider the possibility of having an empty set (no numbers at all). We'll keep track of this as one possible subset.
Next, let's consider each individual number in our list. Each one of them forms a subset by itself.
Now, let's consider all possible pairs of numbers from our list.
After that, we'll consider all possible groups of three numbers from the list.
We continue this process, making sure to try every single combination of numbers, all the way up to using all the numbers in the list together in one big set.
For each of these subsets, we'll carefully check if it is a 'beautiful' subset. This means we'll look at all the pairs of numbers within the subset and confirm that their difference isn't a forbidden number.
Finally, we count all the subsets that meet our 'beautiful' criteria. That number is our answer.

Code Implementation

def the_number_of_beautiful_subsets(numbers, difference):
    number_of_beautiful_subsets = 0
    
    # Iterate through all possible subsets
    for i in range(1 << len(numbers)):
        subset = []
        for j in range(len(numbers)):
            if (i >> j) & 1:
                subset.append(numbers[j])

        # Check if the subset is beautiful
        is_beautiful = True
        #An empty set is defined as beautiful
        if not subset:
            number_of_beautiful_subsets += 1
            continue
        
        for first_index_subset in range(len(subset)):
            for second_index_subset in range(first_index_subset + 1, len(subset)):
                #Check the differences between each pair in the subset
                if abs(subset[first_index_subset] - subset[second_index_subset]) == difference:
                    is_beautiful = False
                    break
            if not is_beautiful:
                break

        # Increment the count if beautiful
        if is_beautiful:
            number_of_beautiful_subsets += 1

    return number_of_beautiful_subsets

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, create a way to remember which numbers are already part of a beautiful subset.
Go through the list of numbers one at a time.
For each number, check if adding it to the current subset would violate the 'beautiful' rule (meaning it shouldn't have a difference of exactly 1 with any number already chosen).
If adding the number is okay, we have two choices: either include it in the current subset, or skip it and leave it out.
Try both possibilities and count how many beautiful subsets can be created in each case.
Combine the counts from both possibilities (including the number and excluding the number) to get the total number of beautiful subsets that can be formed from this point forward.
Use the final count to determine the total number of beautiful subsets for the entire original list of numbers.

Code Implementation

def the_number_of_beautiful_subsets(numbers):
    seen_numbers = set()

    def count_beautiful_subsets(index):
        # Base case: If we've processed all numbers, return 1 (empty set).
        if index == len(numbers):
            return 1

        current_number = numbers[index]
        # Check if adding the number violates the beautiful subset property.
        is_beautiful = True
        if (current_number - 1) in seen_numbers or (current_number + 1) in seen_numbers:
            is_beautiful = False

        count = 0
        # Option 1: Exclude the current number.
        count += count_beautiful_subsets(index + 1)

        # Option 2: Include the current number if it's allowed.
        if is_beautiful:
            seen_numbers.add(current_number)
            count += count_beautiful_subsets(index + 1)

            # Backtrack: Remove the number to explore other possibilities.
            seen_numbers.remove(current_number)

        return count

    # Subtract 1 to exclude the empty subset.
    return count_beautiful_subsets(0) - 1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 1 as an empty set is always a beautiful subset.
Array containing only the number 0	Treat 0 as any other number; if it is the only number, return 2 (empty set and {0})
Array with a single element	Return 2 (empty set and the set containing the single element).
Array with all identical elements	Iterate to check each number's presence and handle the identical number efficiently.
Array containing a wide range of number values	The solution should not overflow when calculating sums of elements, potentially needing 64-bit integers.
Input array with a very large number of elements	Optimize the solution for time complexity, potentially using dynamic programming or memoization to avoid redundant calculations to prevent exceeding time constraints.
Array contains negative numbers or very large positive numbers	Handle negative numbers by not allowing them in the selection or adjust the difference check accordingly, and large numbers to prevent integer overflow during the 'different by one' check.
Input array contains duplicate values	The solution should correctly account for duplicate values when forming subsets, counting each unique set only once, which can be done with the standard iterative subset enumeration logic.

Empty input array

How to Handle:

Return 1 as an empty set is always a beautiful subset.

Array containing only the number 0

How to Handle:

Treat 0 as any other number; if it is the only number, return 2 (empty set and {0})

Array with a single element

How to Handle:

Return 2 (empty set and the set containing the single element).

Array with all identical elements

How to Handle:

Iterate to check each number's presence and handle the identical number efficiently.

Array containing a wide range of number values

How to Handle:

The solution should not overflow when calculating sums of elements, potentially needing 64-bit integers.

Input array with a very large number of elements

How to Handle:

Optimize the solution for time complexity, potentially using dynamic programming or memoization to avoid redundant calculations to prevent exceeding time constraints.

Array contains negative numbers or very large positive numbers

How to Handle:

Handle negative numbers by not allowing them in the selection or adjust the difference check accordingly, and large numbers to prevent integer overflow during the 'different by one' check.

Input array contains duplicate values

How to Handle:

The solution should correctly account for duplicate values when forming subsets, counting each unique set only once, which can be done with the standard iterative subset enumeration logic.