#2178

Medium

auto_awesomeBacktracking search with pruning

LeetCode Problem Workspace

Maximum Split of Positive Even Integers

Determine the largest set of unique positive even integers that sum to a given finalSum using backtracking and greedy selection.

math backtracking greedy

Problem Statement

Given an integer finalSum, split it into the largest possible number of unique positive even integers. If no valid split exists, return an empty list.

Return a list containing the integers of the valid split in any order. For example, finalSum = 12 can be split into [2,4,6] because it maximizes the number of integers, while finalSum = 7 has no valid split, so an empty list is returned.

Examples

Example 1

Input: finalSum = 12

Output: [2,4,6]

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: []

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]

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 Approach

Initial Even Check

First, check if finalSum is divisible by 2. If it is odd, return an empty list immediately since no positive even integer split exists.

Greedy Incremental Selection

Iterate from 2 upwards in steps of 2, adding each even integer to the split if it does not exceed the remaining sum. This maximizes the count of unique integers included in the result.

Backtracking with Pruning

If adding the next even integer would overshoot finalSum, backtrack by adjusting the last added integer or skipping numbers. This ensures all valid combinations are considered efficiently and the maximum number of integers is found.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity varies with finalSum and depends on the number of unique even integers that can be included; greedy selection with pruning minimizes unnecessary iterations. Space complexity depends on the size of the split array and the recursion depth if backtracking is used.

What Interviewers Usually Probe

Ask about handling odd finalSum edge cases.
Probe understanding of greedy selection vs. full backtracking.
Check awareness of pruning to avoid unnecessary recursion.

Common Pitfalls or Variants

Common pitfalls

Forgetting to check if finalSum is odd before starting.
Not enforcing uniqueness of even integers in the split.
Overcomplicating the solution without leveraging greedy incremental selection.

Follow-up variants

Allowing repeated even integers and maximizing count.
Minimizing the largest integer in the split instead of maximizing count.
Splitting into positive odd integers instead of even integers.

FAQ

Can finalSum be negative or zero?

No, finalSum is always at least 1, so only positive splits are possible.

Why do we start adding integers from 2?

Because 2 is the smallest positive even integer and starting here ensures maximum count.

What is the pattern used in Maximum Split of Positive Even Integers?

It primarily uses backtracking search with pruning combined with greedy selection to maximize unique integers.

Is the order of integers in the returned list important?

No, the integers can be returned in any order as long as they form a valid split.

What if finalSum is odd?

There is no valid split for odd finalSum, so the function should return an empty list immediately.

terminal

Solution

Solution 1: Greedy

If $\textit{finalSum}$ is odd, it cannot be split into the sum of several distinct positive even integers, so we directly return an empty array.

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class Solution:
    def maximumEvenSplit(self, finalSum: int) -> List[int]:
        if finalSum & 1:
            return []
        ans = []
        i = 2
        while i <= finalSum:
            finalSum -= i
            ans.append(i)
            i += 2
        ans[-1] += finalSum
        return ans

Continue Practicing

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