# Maximum Alternating Subsequence Sum

## 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.

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  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 <= 105
• 1 <= nums[i] <= 105

### SOLUTION

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]``````