Maximum Split of Positive Even Integers

Problem Description

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

Key Insights

• The sum must consist of unique positive even integers.
• If finalSum is odd, it's impossible to split it into even integers, so the result is an empty list.
• The smallest even integer is 2, and subsequent even integers can be generated by incrementing by 2.
• The approach involves finding the maximum number of unique even integers that can sum up to finalSum.

Space and Time Complexity

Time Complexity: O(sqrt(finalSum)) - since we can iterate through even integers until we reach or exceed finalSum. Space Complexity: O(1) - as we are using a fixed amount of space for variables and the output list.

Solution

The approach to solve this problem is to start from the smallest even integer (2) and keep adding the next even integer (incrementing by 2) until the sum reaches or exceeds finalSum. If the sum exceeds finalSum, we need to adjust the last integer added to ensure the total equals finalSum.

1. Initialize an empty list to store the even integers.
2. Start a loop to add even integers to the list.
3. If at any point the cumulative sum exceeds finalSum, adjust the last integer added to make the total equal to finalSum.
4. Return the list of even integers.

Code Solutions