Maximum Split of Positive Even Integers

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Greedy Algorithms

You are given an integer finalSum. Split it into a sum of a maximum number of unique positive even integers.

For example, given finalSum = 12, the following splits are valid (unique positive even integers summing up to finalSum): (12), (2 + 10), (2 + 4 + 6), and (4 + 8). Among them, (2 + 4 + 6) contains the maximum number of integers. Note that finalSum cannot be split into (2 + 2 + 4 + 4) as all the numbers should be unique.

Return a list of integers that represent a valid split containing a maximum number of integers. If no valid split exists for finalSum, return an empty list. You may return the integers in any order.

Example 1:

Input: finalSum = 12
Output: [2,4,6]
Explanation: The following are valid splits: (12), (2 + 10), (2 + 4 + 6), and (4 + 8).
(2 + 4 + 6) has the maximum number of integers, which is 3. Thus, we return [2,4,6].
Note that [2,6,4], [6,2,4], etc. are also accepted.

Example 2:

Input: finalSum = 7
Output: []
Explanation: There are no valid splits for the given finalSum.
Thus, we return an empty array.

Example 3:

Input: finalSum = 28
Output: [6,8,2,12]
Explanation: The following are valid splits: (2 + 26), (6 + 8 + 2 + 12), and (4 + 24). 
(6 + 8 + 2 + 12) has the maximum number of integers, which is 4. Thus, we return [6,8,2,12].
Note that [10,2,4,12], [6,2,4,16], etc. are also accepted.

Constraints:

1 <= finalSum <= 10¹⁰

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 upper bound on the integer `finalSum`? Is it possible that `finalSum` is zero or negative?
If `finalSum` is odd, should I return an empty list or is there some other expected behavior?
Are there any constraints on the size of the resulting list? For example, is there a maximum number of even integers that can be included in the split?
If there are multiple valid solutions (splits), is there any specific criteria for choosing one (e.g., minimizing the number of integers in the split, maximizing the smallest integer, etc.)?
Is the order of the even integers in the returned list significant, or can they be in any order?

Brute Force Solution

Approach

The brute force approach for splitting even numbers involves trying all possible combinations to find the largest set that sums to the target. We explore every possible breakdown, checking if it works.

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

Start with the smallest possible even number, which is 2. Check if we can include it in our set.
If we include it, subtract it from our target sum and move to the next smallest even number (4).
Keep repeating this process, trying to include each subsequent even number in our set, provided it doesn't exceed the remaining target sum.
If at any point, an even number is too big to include, we skip it and move to the next even number.
We continue this process to explore different branches, where for each even number, we either include it or exclude it, until no target sum remains.
Keep track of all valid sets that we create during this process.
After exploring all possible combinations, choose the set with the largest number of even integers as the answer.

Code Implementation

def maximum_split_brute_force(final_sum: int) -> list[int]:
    if final_sum % 2 != 0:
        return []

    maximum_sets = []

    def backtrack(remaining_sum: int, current_set: list[int], next_even_number: int):
        # If the remaining sum is zero, the current set is valid.
        if remaining_sum == 0:
            maximum_sets.append(current_set[:])
            return

        # If the next even number exceeds the remaining sum, skip it.
        if next_even_number > remaining_sum:
            backtrack(remaining_sum, current_set, next_even_number + 2)
            return

        # Explore the option of not including the current even number.
        backtrack(remaining_sum, current_set, next_even_number + 2)

        # Explore the option of including the current even number.
        current_set.append(next_even_number)
        backtrack(remaining_sum - next_even_number, current_set, next_even_number + 2)
        current_set.pop()

    backtrack(final_sum, [], 2)

    # Find the set with the largest number of elements.
    longest_set = []
    for current_set in maximum_sets:
        if len(current_set) > len(longest_set):
            longest_set = current_set

    return longest_set

Big(O) Analysis

Time Complexity

O(2^(n/2)) – The provided brute force approach involves exploring all possible combinations of even numbers to find the largest set that sums to the target. In the worst case, where the target is a large even number, we are essentially generating a power set of even numbers up to approximately half the target (n/2, where n is the target number). Each even number has the option of being included or excluded in a subset, leading to an exponential number of combinations. Therefore, the time complexity is O(2^(n/2)).

Space Complexity

O(N) – The brute force approach described explores all possible combinations of even numbers. To keep track of all valid sets created during this process, we may need to store multiple sets, potentially up to N/2, where N is the input number (the target sum). Each set stores a varying number of even integers. In the worst-case scenario, we might store a number of sets proportional to N, and each set may also contain a number of elements proportional to N. This gathering of valid sets contributes O(N) space. Therefore the space complexity is O(N).

Optimal Solution

Approach

The best way to split the total amount into even numbers is to start with the smallest even numbers and keep adding them until you can't anymore. If the remaining amount is also even, you're done. If not, adjust the last two numbers to make it work.

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

Start with the smallest even integer, which is 2.
Keep adding the next largest even integer (4, 6, 8, and so on) as long as the total of these numbers is less than or equal to the given amount.
Once you reach a point where adding the next even integer would exceed the given amount, calculate the remainder (the difference between the given amount and the sum of the even integers you've already used).
If the remainder is zero, you have found your solution and can simply return the set of integers used.
If the remainder is not zero, increase the value of the largest used even integer by the amount of the remainder.

Code Implementation

def maximum_split(final_sum: int) -> list[int]:
    result = []
    current_even_number = 2
    current_sum = 0

    while current_sum + current_even_number <= final_sum:
        # Add the current even number to the result list.
        result.append(current_even_number)
        current_sum += current_even_number
        current_even_number += 2

    remainder = final_sum - current_sum

    if remainder > 0:
        # Adjust the largest element if there's a remainder
        result[-1] += remainder

    return result

Big(O) Analysis

Time Complexity

O(sqrt(finalSum)) – The algorithm iteratively adds even numbers (2, 4, 6, ...) until the sum approaches the input finalSum. The number of even integers added, 'k', is related to finalSum by approximately 2 + 4 + 6 + ... + 2k <= finalSum. The left side is 2 * (1 + 2 + 3 + ... + k) = 2 * k * (k+1) / 2 = k * (k+1), which is roughly k^2. Therefore, k^2 is approximately proportional to finalSum, meaning k (the number of iterations) is proportional to the square root of finalSum. Since the number of iterations drives the cost, the time complexity is O(sqrt(finalSum)).

Space Complexity

O(sqrt(finalSum)) – The solution uses a list to store the even integers. In the worst-case scenario, we keep adding even numbers (2, 4, 6, ...) until the next even number would exceed the input finalSum. The number of even integers we add will be proportional to the square root of finalSum, since the sum of the first k even numbers is k*(k+1), which is roughly k^2. Therefore, the space used by the list grows with the square root of finalSum, and the auxiliary space is O(sqrt(finalSum)).

Edge Cases

Case	How to Handle
FinalSum is odd	Return an empty list, as it is impossible to split an odd number into even integers.
FinalSum is 0	Return an empty list, as 0 cannot be represented as a sum of positive even integers.
FinalSum is 2	Return a list containing only 2, as it's the smallest possible even integer.
FinalSum is a large even number that might cause integer overflow if too many small even numbers are added initially	Use a long data type to avoid integer overflow during the summing process.
FinalSum is an even number but impossible to split to more than one even integer (FinalSum < 4)	The while loop condition (sum < finalSum) ensures the algorithm stops when it's impossible to further split.
When the remaining difference can be added to the last element	After the loop, add the remaining difference to the last element in the list instead of appending it to the result.
FinalSum equals to an extremely large number and the algorithm takes too long.	Consider and communicate the time complexity, and explore approaches to optimize if possible, or acknowledge potential performance limitations with very large numbers.
Insufficient memory for very large numbers of split positive even integers	Communicate the space complexity and memory limitations, especially when finalSum is extremely large leading to potentially huge lists.

FinalSum is odd

How to Handle:

Return an empty list, as it is impossible to split an odd number into even integers.

FinalSum is 0

How to Handle:

Return an empty list, as 0 cannot be represented as a sum of positive even integers.

FinalSum is 2

How to Handle:

Return a list containing only 2, as it's the smallest possible even integer.

FinalSum is a large even number that might cause integer overflow if too many small even numbers are added initially

How to Handle:

Use a long data type to avoid integer overflow during the summing process.

FinalSum is an even number but impossible to split to more than one even integer (FinalSum < 4)

How to Handle:

The while loop condition (sum < finalSum) ensures the algorithm stops when it's impossible to further split.

When the remaining difference can be added to the last element

How to Handle:

After the loop, add the remaining difference to the last element in the list instead of appending it to the result.

FinalSum equals to an extremely large number and the algorithm takes too long.

How to Handle:

Consider and communicate the time complexity, and explore approaches to optimize if possible, or acknowledge potential performance limitations with very large numbers.

Insufficient memory for very large numbers of split positive even integers

How to Handle:

Communicate the space complexity and memory limitations, especially when finalSum is extremely large leading to potentially huge lists.