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

impl Solution {
    pub fn maximum_even_split(final_sum: i64) -> Vec<i64> {
        if final_sum % 2 == 1 {
            return vec![];
        }

        let mut final_sum = final_sum;
        let mut 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
    }
}