Formatted question description: https://leetcode.ca/all/1911.html

# 1911. Maximum Alternating Subsequence Sum

## Level

Medium

## Description

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]`

, 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

Use dynamic programming. For `0 <= i < nums.length`

, let `dp[i][0]`

represent the maximum alternating subsequence sum at index `i`

when `nums[i]`

is used as an even index or not used, and `dp[i][1]`

represent the maximum alternating subsequence sum at index `i`

when `nums[i]`

is used as an odd index or not used.

Initially, `dp[0][0] = nums[0]`

and `dp[0][1] = 0`

(since all elements in `nums`

are nonnegative). For `1 <= i < nums.length`

, there exist relations `dp[i][0] = Math.max(Math.max(dp[i - 1][1], 0) + nums[i], dp[i - 1][0])`

and `dp[i][1] = Math.max(Math.max(dp[i - 1][0] - nums[i], 0), dp[i - 1][1])`

.

Finally, return the maximum of `dp[nums.length - 1][0]`

and `dp[nums.length - 1][1]`

.

```
class Solution {
public long maxAlternatingSum(int[] nums) {
int length = nums.length;
long[][] dp = new long[length][2];
dp[0][0] = nums[0];
for (int i = 1; i < length; i++) {
dp[i][0] = Math.max(Math.max(dp[i - 1][1], 0) + nums[i], dp[i - 1][0]);
dp[i][1] = Math.max(Math.max(dp[i - 1][0] - nums[i], 0), dp[i - 1][1]);
}
return Math.max(dp[length - 1][0], dp[length - 1][1]);
}
}
```