**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 [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 <= 10
^{5} - 1 <= nums[i] <= 10
^{5}

**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]
```

*Biweekly Contest 55*

*Biweekly Contest 55*