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Count the Number of K-Free Subsets

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Dynamic Programming

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

A subset of nums is called K-free if it does not contain any two numbers whose sum equals k.

Return the number of K-free subsets of nums.

A subset of nums is an array containing some (possibly none) elements from nums. There should be no duplicate values in the subset.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [2,3,3,4], k = 5
Output: 5
Explanation: The K-free subsets are:
- []
- [2]
- [2, 3]
- [2, 3, 3]
- [4]

Example 2:

Input: nums = [5,5,6], k = 10
Output: 4
Explanation: The K-free subsets are:
- []
- [5]
- [5, 5]
- [6]

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 1000
1 <= k <= 2000

Try coding it in LeetCode

Count the Number of K-Free Subsets

Medium

Asked by:

30 views

Topics:

Dynamic Programming

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

A subset of nums is called K-free if it does not contain any two numbers whose sum equals k.

Return the number of K-free subsets of nums.

A subset of nums is an array containing some (possibly none) elements from nums. There should be no duplicate values in the subset.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [2,3,3,4], k = 5
Output: 5
Explanation: The K-free subsets are:
- []
- [2]
- [2, 3]
- [2, 3, 3]
- [4]

Example 2:

Input: nums = [5,5,6], k = 10
Output: 4
Explanation: The K-free subsets are:
- []
- [5]
- [5, 5]
- [6]

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 1000
1 <= k <= 2000

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 are the constraints on the input array's size (length) and the range of values within the array? Can the array be empty?
Can 'K' be zero or negative? What is the valid range for K?
Are there duplicate numbers in the input array, and if so, how should they be handled when determining if a subset is K-free?
If no K-free subset exists, what should I return (e.g., 0, null, an empty list)?
Is the order of elements within each subset significant? Do subsets {1, 2} and {2, 1} count as the same or different?

Brute Force Solution

Approach

The brute force method for finding K-free subsets involves checking every possible combination of numbers in the given set. We systematically generate each subset and see if it meets the 'K-free' condition, where no two numbers in the subset add up to K.

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

Start by considering an empty subset. This is always a valid K-free subset.
Next, consider subsets containing only one number from the original set. Each of these is a potential K-free subset.
Then, try forming subsets of two numbers. For each pair, check if their sum equals K. If it does, then this subset isn't K-free; otherwise, it is.
Continue creating subsets with three numbers, four numbers, and so on, up to subsets containing all the numbers in the original set.
For each newly created subset, check every possible pair of numbers within the subset to see if their sum equals K. If any pair sums to K, that entire subset is not K-free.
Keep a running count of all the subsets you find that satisfy the K-free condition.
Once you've examined all possible subsets, the final count represents the total number of K-free subsets.

Code Implementation

def count_k_free_subsets_brute_force(numbers, k_value):
    number_of_k_free_subsets = 0
    number_of_elements = len(numbers)

    # Iterate through all possible subsets using bit manipulation.
    for i in range(2**number_of_elements):
        subset = []
        for j in range(number_of_elements):
            # Check if the j-th element is present in the current subset.
            if (i >> j) & 1:
                subset.append(numbers[j])

        is_k_free = True

        # Check if the subset is K-free.
        for first_index in range(len(subset)):
            for second_index in range(first_index + 1, len(subset)):

                # If any two numbers sum to k_value, it's not a K-free subset.
                if subset[first_index] + subset[second_index] == k_value:
                    is_k_free = False
                    break
            if not is_k_free:
                break

        # Increment the count if the subset is K-free.
        if is_k_free:

            #The subset meets the K-free condition.
            number_of_k_free_subsets += 1

    return number_of_k_free_subsets

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm iterates through all possible subsets of the input set. For a set of size n, there are 2^n possible subsets. For each subset, we potentially check all pairs of elements to see if their sum equals K. In the worst case, a subset could have O(n) elements, requiring O(n^2) comparisons. Therefore, the overall time complexity is O(2^n * n^2). However, since 2^n grows faster than n^2, the dominant term is 2^n, and each of those subsets up to the worst case needs to be checked. Therefore, the complexity is approximated as O(n * 2^n).

Space Complexity

O(1) – The brute force method described iterates through all possible subsets. Although generating subsets involves creating temporary subsets, in the context of simply *counting* K-free subsets as described, we only need to keep track of the running count of K-free subsets. This count requires only a single integer variable, and no other auxiliary data structures that scale with the input size (N, the number of elements in the original set) are explicitly mentioned or implied. Therefore, the auxiliary space remains constant, irrespective of the input size.

Optimal Solution

Approach

The problem asks us to count subsets without multiples of a number K. Instead of checking every subset, we group numbers by their remainder when divided by K. The key idea is to then use dynamic programming to count valid subsets based on these remainder groups.

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

First, figure out how many numbers in the given set have each possible remainder when you divide them by K.
Create a table to keep track of the number of valid subsets we've found so far. This table will use the remainders we found earlier.
Start filling in the table. The first entry is about subsets without considering any numbers yet, which always has one possibility (the empty set).
Now, for each remainder, consider whether to include the numbers with that remainder in our subsets or not. If we do, we need to make sure we're not including multiples of K. If we do not, the total count does not change.
Keep updating the table until you've considered all the remainders.
The final answer is in the last entry of your table which gives us the number of K-free subsets.

Code Implementation

def count_k_free_subsets(numbers, k_value):
    remainder_counts = [0] * k_value

    for number in numbers:
        remainder_counts[number % k_value] += 1

    # dp_table[i] is the number of k-free subsets using remainders up to i.
    dp_table = [0] * (k_value + 1)
    dp_table[0] = 1

    for remainder_index in range(1, k_value + 1):
        # If remainder is 0, don't include as they are multiples of k_value.
        if remainder_index == k_value:
            dp_table[remainder_index] = dp_table[remainder_index - 1]

        else:
            # Decide whether to include numbers with this remainder in our subsets.
            dp_table[remainder_index] = (dp_table[remainder_index - 1] *
                                          (1 + remainder_counts[remainder_index - 1]))

    # Subtract 1 to exclude the empty set.
    return dp_table[k_value] - 1

Big(O) Analysis

Time Complexity

O(k + k * k) – The first step involves iterating through the input array of size n to count the occurrences of each remainder when divided by K. This step takes O(n) time. However, since the number of distinct remainders cannot exceed K, this can also be considered as O(k) if we pre-process with a modulo operation. The dynamic programming approach involves iterating through the possible remainders (up to K). Inside the loop there is another loop that considers all possible remainders when making a DP table for counting purposes. The dynamic programming loop contributes O(k * k) to the time complexity since we use the possible remainders as rows and columns of the dp table. Considering all steps, the overall complexity is approximately O(k + k * k), because we can precompute remainders in O(k) before filling the dp table.

Space Complexity

O(K) – The auxiliary space is dominated by the table used for dynamic programming. This table, described in the plain English explanation, has a size proportional to the number of possible remainders when dividing by K. Since there are K possible remainders (0 to K-1), the table requires O(K) space. No other significant data structures are created beyond this table and a few variables, meaning the overall auxiliary space is O(K).

Edge Cases

Case	How to Handle
Null or empty input array	Return 1, representing the empty set as the only K-free subset.
Array with a single element	Return 2, as the single element is K-free along with the empty set.
Large input array size causing potential memory issues	Use dynamic programming with memoization to avoid recalculating subproblems and optimize memory usage.
Array containing all identical values where any two numbers sum to K	Handle this by not picking any elements once K is achievable via summing.
Array with negative numbers	Handle negative numbers appropriately if problem is dealing with sums; ensure valid subset sum calculation.
K is zero	If K is zero, then no two values can sum up to it, and the total number of subsets is 2^n.
Integer overflow when calculating the number of subsets	Use modulo arithmetic during calculation to prevent integer overflow and keep the result within a manageable range.
K is extremely large, potentially larger than the sum of all array elements.	Return 2^n if k is large because no numbers sum to K in this case.

Null or empty input array

How to Handle:

Return 1, representing the empty set as the only K-free subset.

Array with a single element

How to Handle:

Return 2, as the single element is K-free along with the empty set.

Large input array size causing potential memory issues

How to Handle:

Use dynamic programming with memoization to avoid recalculating subproblems and optimize memory usage.

Array containing all identical values where any two numbers sum to K

How to Handle:

Handle this by not picking any elements once K is achievable via summing.

Array with negative numbers

How to Handle:

Handle negative numbers appropriately if problem is dealing with sums; ensure valid subset sum calculation.

K is zero

How to Handle:

If K is zero, then no two values can sum up to it, and the total number of subsets is 2^n.

Integer overflow when calculating the number of subsets

How to Handle:

Use modulo arithmetic during calculation to prevent integer overflow and keep the result within a manageable range.

K is extremely large, potentially larger than the sum of all array elements.

How to Handle:

Return 2^n if k is large because no numbers sum to K in this case.

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 are the constraints on the input array's size (length) and the range of values within the array? Can the array be empty?
Can 'K' be zero or negative? What is the valid range for K?
Are there duplicate numbers in the input array, and if so, how should they be handled when determining if a subset is K-free?
If no K-free subset exists, what should I return (e.g., 0, null, an empty list)?
Is the order of elements within each subset significant? Do subsets {1, 2} and {2, 1} count as the same or different?

Brute Force Solution

Approach

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

Start by considering an empty subset. This is always a valid K-free subset.
Next, consider subsets containing only one number from the original set. Each of these is a potential K-free subset.
Then, try forming subsets of two numbers. For each pair, check if their sum equals K. If it does, then this subset isn't K-free; otherwise, it is.
Continue creating subsets with three numbers, four numbers, and so on, up to subsets containing all the numbers in the original set.
For each newly created subset, check every possible pair of numbers within the subset to see if their sum equals K. If any pair sums to K, that entire subset is not K-free.
Keep a running count of all the subsets you find that satisfy the K-free condition.
Once you've examined all possible subsets, the final count represents the total number of K-free subsets.

Code Implementation

def count_k_free_subsets_brute_force(numbers, k_value):
    number_of_k_free_subsets = 0
    number_of_elements = len(numbers)

    # Iterate through all possible subsets using bit manipulation.
    for i in range(2**number_of_elements):
        subset = []
        for j in range(number_of_elements):
            # Check if the j-th element is present in the current subset.
            if (i >> j) & 1:
                subset.append(numbers[j])

        is_k_free = True

        # Check if the subset is K-free.
        for first_index in range(len(subset)):
            for second_index in range(first_index + 1, len(subset)):

                # If any two numbers sum to k_value, it's not a K-free subset.
                if subset[first_index] + subset[second_index] == k_value:
                    is_k_free = False
                    break
            if not is_k_free:
                break

        # Increment the count if the subset is K-free.
        if is_k_free:

            #The subset meets the K-free condition.
            number_of_k_free_subsets += 1

    return number_of_k_free_subsets

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, figure out how many numbers in the given set have each possible remainder when you divide them by K.
Create a table to keep track of the number of valid subsets we've found so far. This table will use the remainders we found earlier.
Start filling in the table. The first entry is about subsets without considering any numbers yet, which always has one possibility (the empty set).
Now, for each remainder, consider whether to include the numbers with that remainder in our subsets or not. If we do, we need to make sure we're not including multiples of K. If we do not, the total count does not change.
Keep updating the table until you've considered all the remainders.
The final answer is in the last entry of your table which gives us the number of K-free subsets.

Code Implementation

def count_k_free_subsets(numbers, k_value):
    remainder_counts = [0] * k_value

    for number in numbers:
        remainder_counts[number % k_value] += 1

    # dp_table[i] is the number of k-free subsets using remainders up to i.
    dp_table = [0] * (k_value + 1)
    dp_table[0] = 1

    for remainder_index in range(1, k_value + 1):
        # If remainder is 0, don't include as they are multiples of k_value.
        if remainder_index == k_value:
            dp_table[remainder_index] = dp_table[remainder_index - 1]

        else:
            # Decide whether to include numbers with this remainder in our subsets.
            dp_table[remainder_index] = (dp_table[remainder_index - 1] *
                                          (1 + remainder_counts[remainder_index - 1]))

    # Subtract 1 to exclude the empty set.
    return dp_table[k_value] - 1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 1, representing the empty set as the only K-free subset.
Array with a single element	Return 2, as the single element is K-free along with the empty set.
Large input array size causing potential memory issues	Use dynamic programming with memoization to avoid recalculating subproblems and optimize memory usage.
Array containing all identical values where any two numbers sum to K	Handle this by not picking any elements once K is achievable via summing.
Array with negative numbers	Handle negative numbers appropriately if problem is dealing with sums; ensure valid subset sum calculation.
K is zero	If K is zero, then no two values can sum up to it, and the total number of subsets is 2^n.
Integer overflow when calculating the number of subsets	Use modulo arithmetic during calculation to prevent integer overflow and keep the result within a manageable range.
K is extremely large, potentially larger than the sum of all array elements.	Return 2^n if k is large because no numbers sum to K in this case.

Null or empty input array

How to Handle:

Return 1, representing the empty set as the only K-free subset.

Array with a single element

How to Handle:

Return 2, as the single element is K-free along with the empty set.

Large input array size causing potential memory issues

How to Handle:

Use dynamic programming with memoization to avoid recalculating subproblems and optimize memory usage.

Array containing all identical values where any two numbers sum to K

How to Handle:

Handle this by not picking any elements once K is achievable via summing.

Array with negative numbers

How to Handle:

Handle negative numbers appropriately if problem is dealing with sums; ensure valid subset sum calculation.

K is zero

How to Handle:

If K is zero, then no two values can sum up to it, and the total number of subsets is 2^n.

Integer overflow when calculating the number of subsets

How to Handle:

Use modulo arithmetic during calculation to prevent integer overflow and keep the result within a manageable range.

K is extremely large, potentially larger than the sum of all array elements.

How to Handle:

Return 2^n if k is large because no numbers sum to K in this case.