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Find the K-Sum of an Array

Hard

HubSpot

3 views

Topics:

ArraysGreedy AlgorithmsTwo Pointers

You are given an integer array nums and a positive integer k. You can choose any subsequence of the array and sum all of its elements together.

We define the K-Sum of the array as the k^th largest subsequence sum that can be obtained (not necessarily distinct).

Return the K-Sum of the array.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Note that the empty subsequence is considered to have a sum of 0.

Example 1:

Input: nums = [2,4,-2], k = 5
Output: 2
Explanation: All the possible subsequence sums that we can obtain are the following sorted in decreasing order:
- 6, 4, 4, 2, 2, 0, 0, -2.
The 5-Sum of the array is 2.

Example 2:

Input: nums = [1,-2,3,4,-10,12], k = 16
Output: 10
Explanation: The 16-Sum of the array is 10.

Constraints:

n == nums.length
1 <= n <= 10⁵
-10⁹ <= nums[i] <= 10⁹
1 <= k <= min(2000, 2ⁿ)

Try coding it in LeetCode

Find the K-Sum of an Array

Hard

HubSpot

3 views

Topics:

ArraysGreedy AlgorithmsTwo Pointers

You are given an integer array nums and a positive integer k. You can choose any subsequence of the array and sum all of its elements together.

We define the K-Sum of the array as the k^th largest subsequence sum that can be obtained (not necessarily distinct).

Return the K-Sum of the array.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Note that the empty subsequence is considered to have a sum of 0.

Example 1:

Input: nums = [2,4,-2], k = 5
Output: 2
Explanation: All the possible subsequence sums that we can obtain are the following sorted in decreasing order:
- 6, 4, 4, 2, 2, 0, 0, -2.
The 5-Sum of the array is 2.

Example 2:

Input: nums = [1,-2,3,4,-10,12], k = 16
Output: 10
Explanation: The 16-Sum of the array is 10.

Constraints:

n == nums.length
1 <= n <= 10⁵
-10⁹ <= nums[i] <= 10⁹
1 <= k <= min(2000, 2ⁿ)

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 size of the input array `nums` and the value of `k`? Specifically, what are the maximum values for these?
Can the input array `nums` contain negative numbers, zeros, or floating-point numbers, or is it strictly positive integers?
If `k` is greater than the number of positive elements in the sorted array, what should I do? Should I treat the excess as if it were zero, or is this an invalid input?
What should I return if the input array `nums` is null or empty?
Is the input array `nums` guaranteed to be sortable (e.g., does it contain objects that lack a defined ordering)? Assume the input contains only integers.

Brute Force Solution

Approach

The brute force strategy for finding the K-Sum involves checking every possible combination of numbers in the given set. This means looking at all possible groups of 'K' numbers and calculating their sum. We then see if any of those sums match our target value.

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

First, consider every possible group of 'K' numbers from the set.
Calculate the sum of the numbers in each of these groups.
For each sum, compare it to the target value we are looking for.
If any of the sums matches the target value, then we have found a solution.
Continue this process until all possible groups of 'K' numbers have been checked.
If no sum matches the target value after checking every group, then there is no solution.

Code Implementation

def k_sum_brute_force(numbers, k_value, target):    number_of_numbers = len(numbers)
    if k_value <= 0 or k_value > number_of_numbers:
        return False

    # Generate all possible combinations of k_value numbers
    def find_combinations(current_combination, start_index):
        if len(current_combination) == k_value:
            combination_sum = sum(current_combination)
            if combination_sum == target:
                return True
            else:
                return False

        # Optimization: Avoid unnecessary exploration if remaining elements are insufficient
        if start_index >= number_of_numbers:
            return False

        # Explore including the current number
        if find_combinations(current_combination + [numbers[start_index]], start_index + 1):
            return True
        
        # Explore excluding the current number
        if find_combinations(current_combination, start_index + 1):
            return True

        return False

    # Initiate search for combinations
    return find_combinations([], 0)

Big(O) Analysis

Time Complexity

O(n^K) – The brute force approach considers all possible combinations of K numbers from an array of size n. The number of ways to choose K elements from n is given by the binomial coefficient 'n choose K', which is n! / (K! * (n-K)!). In the worst case, we might have to examine all these combinations. The cost of generating each combination and summing its elements is O(K). However, generating all combinations dominates the time complexity. Therefore, the overall time complexity is O(n^K) as K approaches n, because it represents the leading term in calculating the number of combinations.

Space Complexity

O(1) – The provided plain English explanation outlines a brute-force approach that iterates through all possible K-sized combinations within an array of size N. It doesn't explicitly mention any auxiliary data structures like temporary lists, hash maps, or recursion. Only the sum for each combination and the target value need to be stored for comparison, requiring a constant amount of extra space. Therefore, the auxiliary space complexity is O(1), as it remains constant irrespective of the input array's size.

Optimal Solution

Approach

The optimal solution first identifies if the desired sum is even possible. Then, it uses a divide-and-conquer strategy, breaking down the problem into smaller, more manageable subproblems that are easier to solve.

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

Check if the sum of all the positive numbers in the input is less than the target sum. If it is, there's no solution because we can't reach the target.
Also check if the sum of all the negative numbers in the input is greater than the target sum. If it is, there's no solution because we can't reach the target.
Sort the numbers from smallest to largest. This makes it easier to find combinations that add up to the desired sum.
Create a function to solve the general problem of finding 'N' numbers that sum to a target value. This function will call itself with smaller versions of the problem.
Start with the biggest value of N (in this case, K). The function looks for combinations of 'K' numbers that add up to the target.
If 'K' is 2, then use the 'two-pointer' technique, where you have one pointer at the start of the list and one at the end. Move them closer to each other based on whether the current sum is too high or too low. If the current sum equals the target value, we found the solution.
If 'K' is greater than 2, fix one number in the list and then call the function again to find 'K-1' numbers that add up to the target minus the number you fixed.
Make sure to skip over numbers that are the same to avoid duplicate solutions. This is especially important after you fix a number.
Continue this process until you find all the combinations of 'K' numbers that add up to the target, or until you've exhausted all possibilities.

Code Implementation

def find_k_sum(numbers, k_value, target_sum):
    positive_sum = sum(number for number in numbers if number > 0)
    negative_sum = sum(number for number in numbers if number < 0)

    if positive_sum < target_sum or negative_sum > target_sum:
        return []

    numbers.sort()

    def find_n_sum(numbers, number_count, target_sum):
        results = []

        if number_count == 2:
            left_index = 0
            right_index = len(numbers) - 1

            while left_index < right_index:
                current_sum = numbers[left_index] + numbers[right_index]

                if current_sum == target_sum:
                    results.append([numbers[left_index], numbers[right_index]])
                    left_index += 1
                    right_index -= 1
                    while left_index < right_index and numbers[left_index] == numbers[left_index - 1]:
                        left_index += 1
                    while left_index < right_index and numbers[right_index] == numbers[right_index + 1]:
                        right_index -= 1
                elif current_sum < target_sum:
                    left_index += 1
                else:
                    right_index -= 1
        else:
            for index in range(len(numbers) - number_count + 1):
                # Avoid duplicate solutions by skipping same values
                if index > 0 and numbers[index] == numbers[index - 1]:
                    continue

                remaining_numbers = numbers[index + 1:]
                sub_target = target_sum - numbers[index]
                sub_results = find_n_sum(remaining_numbers, number_count - 1, sub_target)

                for sub_result in sub_results:
                    results.append([numbers[index]] + sub_result)

        return results

    # Initiate the process to find K numbers
    return find_n_sum(numbers, k_value, target_sum)

Big(O) Analysis

Time Complexity

O(n^(k-1)) – Sorting the array takes O(n log n) time, which is dominated by other operations if k > 2. The kSum function uses a recursive approach. When k equals 2, the two-pointer technique is applied, taking O(n) time. For each k greater than 2, the algorithm iterates through the array (n elements) and recursively calls kSum with k-1. This results in a nested loop structure that goes k-2 levels deep, so the overall complexity becomes O(n^(k-1)). Therefore, the dominant factor determining runtime is this recursive exploration of combinations, making the algorithm O(n^(k-1)).

Space Complexity

O(N) – The space complexity is dominated by the recursion depth of the kSum function. In the worst-case scenario, the kSum function can be called recursively up to N times (where N is the number of elements in the input array) when k is reduced to 2. Each recursive call adds a new frame to the call stack, thus the auxiliary space used grows proportionally with the input size N. Additionally, the sorting step might take O(N) space depending on the sorting algorithm used, but the recursive call stack depth is the dominating factor in space complexity.

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 immediately since there are no elements to sum.
k is 0 (no sign changes)	Calculate and return the sum of positive elements directly without changing signs.
k is greater than the number of positive elements	Change the sign of all positive elements, and then change the sign of the largest (k - numPositive) negative elements to positive.
All numbers in the array are negative or zero	Return 0, as there are no positive numbers to make negative.
Array contains very large positive numbers leading to potential integer overflow during summation	Use a larger data type (e.g., long) to store the sums and handle potentially overflowing calculations.
Array contains duplicate positive values; which ones do we negate?	Sort the array and negate the first k positive elements as per the definition, addressing duplicates by their sorted order.
Array contains a mix of small and extremely large numbers, potentially impacting sorting performance	Utilize an efficient sorting algorithm with good average-case performance (e.g., merge sort or quicksort) to handle varied value ranges.
k is a negative number	Return 0 or throw an IllegalArgumentException as negative k values are not defined in this scenario

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 size of the input array `nums` and the value of `k`? Specifically, what are the maximum values for these?
Can the input array `nums` contain negative numbers, zeros, or floating-point numbers, or is it strictly positive integers?
If `k` is greater than the number of positive elements in the sorted array, what should I do? Should I treat the excess as if it were zero, or is this an invalid input?
What should I return if the input array `nums` is null or empty?
Is the input array `nums` guaranteed to be sortable (e.g., does it contain objects that lack a defined ordering)? Assume the input contains only integers.

Brute Force Solution

Approach

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

First, consider every possible group of 'K' numbers from the set.
Calculate the sum of the numbers in each of these groups.
For each sum, compare it to the target value we are looking for.
If any of the sums matches the target value, then we have found a solution.
Continue this process until all possible groups of 'K' numbers have been checked.
If no sum matches the target value after checking every group, then there is no solution.

Code Implementation

def k_sum_brute_force(numbers, k_value, target):    number_of_numbers = len(numbers)
    if k_value <= 0 or k_value > number_of_numbers:
        return False

    # Generate all possible combinations of k_value numbers
    def find_combinations(current_combination, start_index):
        if len(current_combination) == k_value:
            combination_sum = sum(current_combination)
            if combination_sum == target:
                return True
            else:
                return False

        # Optimization: Avoid unnecessary exploration if remaining elements are insufficient
        if start_index >= number_of_numbers:
            return False

        # Explore including the current number
        if find_combinations(current_combination + [numbers[start_index]], start_index + 1):
            return True
        
        # Explore excluding the current number
        if find_combinations(current_combination, start_index + 1):
            return True

        return False

    # Initiate search for combinations
    return find_combinations([], 0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Check if the sum of all the positive numbers in the input is less than the target sum. If it is, there's no solution because we can't reach the target.
Also check if the sum of all the negative numbers in the input is greater than the target sum. If it is, there's no solution because we can't reach the target.
Sort the numbers from smallest to largest. This makes it easier to find combinations that add up to the desired sum.
Create a function to solve the general problem of finding 'N' numbers that sum to a target value. This function will call itself with smaller versions of the problem.
Start with the biggest value of N (in this case, K). The function looks for combinations of 'K' numbers that add up to the target.
If 'K' is 2, then use the 'two-pointer' technique, where you have one pointer at the start of the list and one at the end. Move them closer to each other based on whether the current sum is too high or too low. If the current sum equals the target value, we found the solution.
If 'K' is greater than 2, fix one number in the list and then call the function again to find 'K-1' numbers that add up to the target minus the number you fixed.
Make sure to skip over numbers that are the same to avoid duplicate solutions. This is especially important after you fix a number.
Continue this process until you find all the combinations of 'K' numbers that add up to the target, or until you've exhausted all possibilities.

Code Implementation

def find_k_sum(numbers, k_value, target_sum):
    positive_sum = sum(number for number in numbers if number > 0)
    negative_sum = sum(number for number in numbers if number < 0)

    if positive_sum < target_sum or negative_sum > target_sum:
        return []

    numbers.sort()

    def find_n_sum(numbers, number_count, target_sum):
        results = []

        if number_count == 2:
            left_index = 0
            right_index = len(numbers) - 1

            while left_index < right_index:
                current_sum = numbers[left_index] + numbers[right_index]

                if current_sum == target_sum:
                    results.append([numbers[left_index], numbers[right_index]])
                    left_index += 1
                    right_index -= 1
                    while left_index < right_index and numbers[left_index] == numbers[left_index - 1]:
                        left_index += 1
                    while left_index < right_index and numbers[right_index] == numbers[right_index + 1]:
                        right_index -= 1
                elif current_sum < target_sum:
                    left_index += 1
                else:
                    right_index -= 1
        else:
            for index in range(len(numbers) - number_count + 1):
                # Avoid duplicate solutions by skipping same values
                if index > 0 and numbers[index] == numbers[index - 1]:
                    continue

                remaining_numbers = numbers[index + 1:]
                sub_target = target_sum - numbers[index]
                sub_results = find_n_sum(remaining_numbers, number_count - 1, sub_target)

                for sub_result in sub_results:
                    results.append([numbers[index]] + sub_result)

        return results

    # Initiate the process to find K numbers
    return find_n_sum(numbers, k_value, target_sum)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 immediately since there are no elements to sum.
k is 0 (no sign changes)	Calculate and return the sum of positive elements directly without changing signs.
k is greater than the number of positive elements	Change the sign of all positive elements, and then change the sign of the largest (k - numPositive) negative elements to positive.
All numbers in the array are negative or zero	Return 0, as there are no positive numbers to make negative.
Array contains very large positive numbers leading to potential integer overflow during summation	Use a larger data type (e.g., long) to store the sums and handle potentially overflowing calculations.
Array contains duplicate positive values; which ones do we negate?	Sort the array and negate the first k positive elements as per the definition, addressing duplicates by their sorted order.
Array contains a mix of small and extremely large numbers, potentially impacting sorting performance	Utilize an efficient sorting algorithm with good average-case performance (e.g., merge sort or quicksort) to handle varied value ranges.
k is a negative number	Return 0 or throw an IllegalArgumentException as negative k values are not defined in this scenario