# Solving Leetcode Interviews in Seconds with AI: Maximum Alternating Subsequence Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1911" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 

	# Explanation
	Here's a breakdown of the optimal approach, complexity analysis, and the Python code:

*   **Dynamic Programming with Two States:** The core idea is to use dynamic programming to keep track of two states: `even` and `odd`. `even` represents the maximum alternating sum ending with an element at an even index, and `odd` represents the maximum alternating sum ending with an element at an odd index.
*   **Iterative Update:** Iterate through the input array `nums`. At each element, update `even` and `odd` based on the current element. Either include the element in the subsequence (and update the sums accordingly), or exclude it (and keep the previous maximums).
*   **Maximization:** The final answer will be the `even` state after iterating through the entire array, as we want the subsequence to end with an element conceptually considered at an even position.

*   **Runtime Complexity:** O(n), where n is the length of the input array `nums`.
*   **Storage Complexity:** O(1) (constant space), as we are only using two variables to store the `even` and `odd` states.

	
	# Code
	```python
	def maxAlternatingSum(nums):
    """
    Calculates the maximum alternating sum of any subsequence of nums.

    Args:
        nums: A list of integers.

    Returns:
        The maximum alternating sum.
    """
    even = 0
    odd = 0

    for num in nums:
        even = max(even, odd + num)
        odd = max(odd, even - num)

    return even
	```