2178 - Maximum Split of Positive Even Integers (Medium)

Problem Link

https://leetcode.com/problems/maximum-split-of-positive-even-integers/

Problem Statement

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): (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 some valid splits: (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 some 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 <= 10^10

Approach 1: Greedy

If $finalSum$ is odd, then return empty array.

Otherwise, we start from the lowest even number $x = 2$. Then add it to our final array. Subtract $finalSum$ from $x$ and set $x$ to the next even number. Do the same logic until $x$ cannot be used. At the end, add the remainder to the last element of the array.

Written by @wingkwong

class Solution {
public:
    vector<long long> maximumEvenSplit(long long finalSum) {
        vector<long long> ans;
        // final sum is odd, return empty array
        if (finalSum & 1) return ans;
        // add lowest even number
        long long x = 2;
        // check if we can use x
        while (x <= finalSum) {
            // add it to answer
            ans.push_back(x);
            // update finalSum and set x to the next even number
            finalSum -= x, x += 2;
        }
        // add the remainder to the last element
        ans.back() += finalSum;
        return ans;
    }
};