# 1911. Maximum Alternating Subsequence Sum

## 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] (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 <= 105
• 1 <= nums[i] <= 105

## Solutions

• class Solution {
public long maxAlternatingSum(int[] nums) {
int n = nums.length;
long[] f = new long[n + 1];
long[] g = new long[n + 1];
for (int i = 1; i <= n; ++i) {
f[i] = Math.max(g[i - 1] - nums[i - 1], f[i - 1]);
g[i] = Math.max(f[i - 1] + nums[i - 1], g[i - 1]);
}
return Math.max(f[n], g[n]);
}
}

• class Solution {
public:
long long maxAlternatingSum(vector<int>& nums) {
int n = nums.size();
vector<long long> f(n + 1), g(n + 1);
for (int i = 1; i <= n; ++i) {
f[i] = max(g[i - 1] - nums[i - 1], f[i - 1]);
g[i] = max(f[i - 1] + nums[i - 1], g[i - 1]);
}
return max(f[n], g[n]);
}
};

• class Solution:
def maxAlternatingSum(self, nums: List[int]) -> int:
n = len(nums)
f = [0] * (n + 1)
g = [0] * (n + 1)
for i, x in enumerate(nums, 1):
f[i] = max(g[i - 1] - x, f[i - 1])
g[i] = max(f[i - 1] + x, g[i - 1])
return max(f[n], g[n])


• func maxAlternatingSum(nums []int) int64 {
n := len(nums)
f := make([]int, n+1)
g := make([]int, n+1)
for i, x := range nums {
i++
f[i] = max(g[i-1]-x, f[i-1])
g[i] = max(f[i-1]+x, g[i-1])
}
return int64(max(f[n], g[n]))
}

• function maxAlternatingSum(nums: number[]): number {
const n = nums.length;
const f: number[] = new Array(n + 1).fill(0);
const g = f.slice();
for (let i = 1; i <= n; ++i) {
f[i] = Math.max(g[i - 1] + nums[i - 1], f[i - 1]);
g[i] = Math.max(f[i - 1] - nums[i - 1], g[i - 1]);
}
return Math.max(f[n], g[n]);
}