<p>The <strong>alternating sum</strong> of a <strong>0-indexed</strong> array is defined as the <strong>sum</strong> of the elements at <strong>even</strong> indices <strong>minus</strong> the <strong>sum</strong> of the elements at <strong>odd</strong> indices.</p>

<ul>
	<li>For example, the alternating sum of <code>[4,2,5,3]</code> is <code>(4 + 5) - (2 + 3) = 4</code>.</li>
</ul>

<p>Given an array <code>nums</code>, return <em>the <strong>maximum alternating sum</strong> of any subsequence of </em><code>nums</code><em> (after <strong>reindexing</strong> the elements of the subsequence)</em>.</p>

<ul>
</ul>

<p>A <strong>subsequence</strong> of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements&#39; relative order. For example, <code>[2,7,4]</code> is a subsequence of <code>[4,<u>2</u>,3,<u>7</u>,2,1,<u>4</u>]</code> (the underlined elements), while <code>[2,4,2]</code> is not.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u>4</u>,<u>2</u>,<u>5</u>,3]
<strong>Output:</strong> 7
<strong>Explanation:</strong> It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) - 2 = 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,6,7,<u>8</u>]
<strong>Output:</strong> 8
<strong>Explanation:</strong> It is optimal to choose the subsequence [8] with alternating sum 8.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [<u>6</u>,2,<u>1</u>,2,4,<u>5</u>]
<strong>Output:</strong> 10
<strong>Explanation:</strong> It is optimal to choose the subsequence [6,1,5] with alternating sum (6 + 5) - 1 = 10.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
        # return recursion(nums)
        return memo(nums)
def recursion(nums):
    res = 0
    n = len(nums)
    dp = {}
    
    def dfs(idx, flag):
        # base case
        if idx >= n:
            return 0
        if (idx,flag) in dp:
            return dp[(idx,flag)]
        # take
        if flag == 'even':
            take = nums[idx] + dfs(idx+1, 'odd')
        else:
            take = -nums[idx] + dfs(idx+1, 'even')
        not_take = 0 + dfs(idx+1, flag)
        
        res = max(take, not_take)
        dp[(idx,flag)] = res
        return res

    return dfs(0, 'even')

def memo(nums):
    n= len(nums)
    dp = {(0,0):0, (0,1):0} # if there is no num then there is no max subsequence sum for both even and odd cases 
    # define states
    for i in range(1, n+1):
        dp[(i,0)] = max(dp[(i-1,1)] - nums[i-1], dp[(i-1,0)])
        dp[(i,1)] = max(dp[(i-1,0)] + nums[i-1], dp[(i-1,1)])
    return max(dp[(n,0)], dp[(n,1)])
    
1274