README.md

2178. 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

Solutions (Rust)

1. Solution

implSolution{pubfnmaximum_even_split(final_sum:i64) -> Vec<i64>{if final_sum % 2 == 1{returnvec![];}letmut final_sum = final_sum;letmut ret = vec![];for x in(2..=final_sum).step_by(2){if final_sum - x <= x { ret.push(final_sum);break;} ret.push(x); final_sum -= x;} ret }}