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2178. Maximum Split of Positive Even Integers

Description

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

  • class Solution {
    public List<Long> maximumEvenSplit(long finalSum) {
        List<Long> ans = new ArrayList<>();
        if (finalSum % 2 == 1) {
            return ans;
        }
        for (long i = 2; i <= finalSum; i += 2) {
            ans.add(i);
            finalSum -= i;
        }
        ans.add(ans.remove(ans.size() - 1) + finalSum);
        return ans;
    }
}

  • class Solution {
public:
    vector<long long> maximumEvenSplit(long long finalSum) {
        vector<long long> ans;
        if (finalSum % 2) return ans;
        for (long long i = 2; i <= finalSum; i += 2) {
            ans.push_back(i);
            finalSum -= i;
        }
        ans.back() += finalSum;
        return ans;
    }
};

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


  • func maximumEvenSplit(finalSum int64) (ans []int64) {
	if finalSum%2 == 1 {
		return
	}
	for i := int64(2); i <= finalSum; i += 2 {
		ans = append(ans, i)
		finalSum -= i
	}
	ans[len(ans)-1] += finalSum
	return
}

  • function maximumEvenSplit(finalSum: number): number[] {
    const ans: number[] = [];
    if (finalSum % 2 === 1) {
        return ans;
    }
    for (let i = 2; i <= finalSum; i += 2) {
        ans.push(i);
        finalSum -= i;
    }
    ans[ans.length - 1] += finalSum;
    return ans;
}


  • public class Solution {
    public IList<long> MaximumEvenSplit(long finalSum) {
        IList<long> ans = new List<long>();
        if (finalSum % 2 == 1) {
            return ans;
        }
        for (long i = 2; i <= finalSum; i += 2) {
            ans.Add(i);
            finalSum -= i;
        }
        ans[ans.Count - 1] += finalSum;
        return ans;
    }
}

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