LeetCode in Kotlin
2178. Maximum Split of Positive Even Integers
Medium
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 tofinalSum):(12),(2 + 10),(2 + 4 + 6), and(4 + 8). Among them,(2 + 4 + 6)contains the maximum number of integers. Note thatfinalSumcannot 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
Solution
class Solution {
fun maximumEvenSplit(finalSum: Long): List<Long> {
var curr: Long = 2
var remainingSum = finalSum
val result: MutableList<Long> = ArrayList()
if (finalSum % 2 != 0L) {
return result
}
while (remainingSum >= curr) {
result.add(curr)
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]
*/
val lastSum = result[result.size - 1]
result.removeAt(result.size - 1)
result.add(lastSum + remainingSum)
return result
}
}