Maximum Alternating Subsequence Sum Solution
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.
Input: nums = [4,2,5,3]
Explanation: It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) – 2 = 7.
Input: nums = [5,6,7,8]
Explanation: It is optimal to choose the subsequence  with alternating sum 8.
Input: nums = [6,2,1,2,4,5]
Explanation: It is optimal to choose the subsequence [6,1,5] with alternating sum (6 + 5) – 1 = 10.
- 1 <= nums.length <= 105
- 1 <= nums[i] <= 105
Program: Maximum Alternating Subsequence Sum Solution in Python
class Solution: def maxAlternatingSum(self, nums: List[int]) -> int: ans = 0 direction = 'down' n = len(nums) for i in range(n-1): if direction == 'down' and nums[i] >= nums[i+1]: ans += nums[i] direction = 'up' elif direction == 'up' and nums[i] <= nums[i+1]: ans -= nums[i] direction = 'down' if direction == 'up': return ans return ans + nums[-1]