#+ANKI_DECK: study_deck_02
* TODO 1911. Maximum Alternating Subsequence Sum :medium:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*1911. Maximum Alternating Subsequence Sum][1911. Maximum Alternating Subsequence Sum]]
:END:

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:*


#+begin_src
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.
#+end_src


*Example 2:*


#+begin_src
Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subsequence [8] with alternating sum 8.
#+end_src


*Example 3:*


#+begin_src
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.
#+end_src


*Constraints:*

 - ~1 <= nums.length <= 10^{5}~

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

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python :lc-problem 1911 :lc-lang python3
class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
#+end_src

** TODO C++
#+begin_src cpp :lc-problem 1911
class Solution {
public:
    long long maxAlternatingSum(vector<int>& nums) {
        
    }
};
#+end_src