Example
1911. Maximum Alternating Subsequence Sum
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.
Answers
Dynamic Programming with space compression optimization
var maxAlternatingSum = function(nums) {
let even = nums[0], odd = 0
for (const num of nums) {
const t = even
even = Math.max(even, odd + num)
odd = Math.max(odd, t - num)
}
return Math.max(even, odd)
};func maxAlternatingSum(nums []int) int64 {
even, odd := int64(nums[0]), int64(0)
for _, num := range nums {
t := even
even = max(even, odd + int64(num))
odd = max(odd, t - int64(num))
}
return max(even, odd)
}
func max(a, b int64) int64 {
if a > b {
return a
}
return b
}class Solution {
function maxAlternatingSum($nums) {
$even = $nums[0];
$odd = 0;
foreach ($nums as $num) {
$t = $even;
$even = max($even, $odd + $num);
$odd = max($odd, $even - $num);
}
return max($even, $odd);
}
}class Solution {
public long maxAlternatingSum(int[] nums) {
long even = nums[0], odd = 0;
for (int num : nums) {
long t = even;
even = Math.max(even, odd + num);
odd = Math.max(odd, t - num);
}
return Math.max(even, odd);
}
}public class Solution {
public long MaxAlternatingSum(int[] nums) {
long even = nums[0], odd = 0;
foreach (int num in nums) {
long t = even;
even = Math.Max(even, odd + num);
odd = Math.Max(odd, t - num);
}
return Math.Max(even, odd);
}
}class Solution {
public:
long long maxAlternatingSum(vector<int>& nums) {
long long even = nums[0], odd = 0;
for (int num : nums) {
long long t = even;
even = max(even, odd + num);
odd = max(odd, even - num);
}
return max(even, odd);
}
};#define MAX(a, b) (a > b ? a : b)
long long maxAlternatingSum(int* nums, int numsSize){
long long even = nums[0], odd = 0;
for (int i = 0; i < numsSize; i++) {
long long t = even;
even = MAX(even, odd + nums[i]);
odd = MAX(odd, even - nums[i]);
}
return MAX(even, odd);
}class Solution:
def maxAlternatingSum(self, nums: List[int]) -> int:
even, odd = nums[0], 0
for num in nums:
t = even
even, odd = max(even, odd + num), max(odd, t - num)
return max(even, odd)