Medium
You are given an integer finalSum
. Split it into a sum of a maximum number of unique positive even integers.
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
import java.util.ArrayList;
import java.util.List;
public class Solution {
public List<Long> maximumEvenSplit(long finalSum) {
long curr = 2;
long remainingSum = finalSum;
List<Long> result = new ArrayList<>();
if (finalSum % 2 != 0) {
return result;
}
while (remainingSum >= curr) {
result.add(curr);
remainingSum = remainingSum - curr;
curr += 2;
}
/*
go greedily by starting from smallest even number
for target = 16 after the while loop
remainingSum = 4
curr = 8 (if we add 8 it exceeds the target 16)
result = [2,4,6]
so remove 6 from list and add it to remainigSum and insert to list
result = [2,4,10]
*/
long lastSum = result.get(result.size() - 1);
result.remove(result.size() - 1);
result.add(lastSum + remainingSum);
return result;
}
}