{x}
blog image

Maximum Split of Positive Even Integers

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

Solution Explanation for Maximum Split of Positive Even Integers

This problem asks to split a given positive integer finalSum into a maximum number of unique positive even integers. The solution employs a greedy approach.

Approach:

  1. Handle Odd Numbers: If finalSum is odd, no solution exists because the sum of even numbers is always even. The function immediately returns an empty list.

  2. Greedy Splitting: If finalSum is even, the algorithm iteratively subtracts the smallest available even number (starting from 2, then 4, 6, and so on) from finalSum. Each subtracted number is added to the result list. This continues until the remaining finalSum is either zero or smaller than the next even number to be considered.

  3. Handling Remainder: The remaining finalSum (if non-zero) is added to the last element in the result list. This ensures all the even numbers are unique and the sum equals the original finalSum.

Time Complexity Analysis:

The loop iterates until i exceeds finalSum. In the worst case, i will reach approximately √finalSum. Therefore, the time complexity is O(√finalSum). This is because the sum of even numbers up to 2k is k(k+1), and we're essentially finding the largest k such that k(k+1) ≤ finalSum. Solving this quadratic equation gives a complexity proportional to the square root of finalSum.

Space Complexity Analysis:

The space complexity is dominated by the size of the ans (result) list. In the worst-case scenario, the list can contain approximately √finalSum elements. Thus, the space complexity is O(√finalSum). We can consider this to be linear in the output size if the output is considered significant.

Code Examples (Python):

class Solution:
    def maximumEvenSplit(self, finalSum: int) -> List[int]:
        if finalSum % 2:
            return []
        ans = []
        i = 2
        while i <= finalSum:
            finalSum -= i
            ans.append(i)
            i += 2
        ans[-1] += finalSum  # Add remainder to the last element
        return ans

The other language examples follow the same logic, differing only in syntax. The key is the greedy iterative subtraction and the final adjustment to include any remaining sum.