51. 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.

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.

Solution: (Finding local maxima and minima)

class Solution
{
public:
    long long maxAlternatingSum(vector<int> &nums)
    {

        int n = nums.size();
        long long int s = 0;

        int i = 0;

        while (i < n)
        {

            //Local maximun
            while (i < n - 1 && nums[i] <= nums[i + 1])
            {
                i++;
            }

            if (i == n - 1)
            {
                s += nums[i];
                break;
            }
            else
            {
                s += nums[i];
            }
            
            //Local minimum
            while (i < n - 1 && nums[i] >= nums[i + 1])
            {
                i++;
            }

            if (i <= n - 1)
            {
                s -= nums[i];
            }
        }

        return s;
    }
};

Time Complexity: O(n)