LeetCode in Kotlin
1911. Maximum Alternating Subsequence Sum
Medium
The alternating sum of a 0-indexed array is defined as the sum of the elements at even indices minus the sum of the elements at odd indices.
- For example, the alternating sum of
[4,2,5,3]is(4 + 5) - (2 + 3) = 4.
Given an array nums, return the maximum alternating sum of any subsequence of nums (after reindexing the elements of the subsequence).
A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements’ relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.
Example 1:
Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) - 2 = 7.
Example 2:
Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subsequence [8] with alternating sum 8.
Example 3:
Input: nums = [6,2,1,2,4,5]
Output: 10
Explanation: It is optimal to choose the subsequence [6,1,5] with alternating sum (6 + 5) - 1 = 10.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105
Solution
class Solution {
fun maxAlternatingSum(nums: IntArray): Long {
val n = nums.size
var even = nums[0].toLong()
var odd: Long = 0
for (i in 1 until n) {
even = Math.max(even, Math.max(odd + nums[i], nums[i].toLong()))
odd = Math.max(odd, Math.max(even - nums[i], 0))
}
return Math.max(even, odd)
}
}