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.
<li>For example, the alternating sum of <code>[4,2,5,3]</code> is <code>(4 + 5) - (2 + 3) = 4</code>.</li>
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:
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>
Solutions
Solution 1
PythonJavaC++GoTypeScript
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])(code-box)
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]);
}
}(code-box)
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]);
}
};(code-box)
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]))
}(code-box)
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]);
}(code-box)
Solution 2
PythonJavaC++GoTypeScript
class Solution:
def maxAlternatingSum(self, nums: List[int]) -> int:
f = g = 0
for x in nums:
f, g = max(g - x, f), max(f + x, g)
return max(f, g)(code-box)
class Solution {
public long maxAlternatingSum(int[] nums) {
long f = 0, g = 0;
for (int x : nums) {
long ff = Math.max(g - x, f);
long gg = Math.max(f + x, g);
f = ff;
g = gg;
}
return Math.max(f, g);
}
}(code-box)
class Solution {
public:
long long maxAlternatingSum(vector<int>& nums) {
long long f = 0, g = 0;
for (int& x : nums) {
long ff = max(g - x, f), gg = max(f + x, g);
f = ff, g = gg;
}
return max(f, g);
}
};(code-box)
func maxAlternatingSum(nums []int) int64 {
var f, g int
for _, x := range nums {
f, g = max(g-x, f), max(f+x, g)
}
return int64(max(f, g))
}(code-box)
function maxAlternatingSum(nums: number[]): number {
let [f, g] = [0, 0];
for (const x of nums) {
[f, g] = [Math.max(g - x, f), Math.max(f + x, g)];
}
return g;
}(code-box)
