<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:
        n = len(nums)   
        dp = [[0,0] for _ in range(n)] # initialize dp　=> 2D DP の初期化までは考え方が合ってた
        dp[0][0] = nums[0] # 最初の値が偶数
        dp[0][1] = 0 # 最初の値が奇数

        # 一個前の状態から算出できた　 => current, prev パターンだと思った => 時間計算量が間に合わない
        for i in range(1, n): # iterate through nums starting from index 1
            dp[i][0] = max(nums[i] + dp[i-1][1], dp[i-1][0]) # i の要素を偶数として追加する場合と追加しない場合で最大値を比べる
            dp[i][1] = max(-nums[i] + dp[i-1][0], dp[i-1][1]) # i の要素を奇数として追加する場合と追加しない場合で最大値を比べる
        
        return max(dp[-1]) # find the maximum of the last array of dp of whether the last value is plus or minus, this will be our answer.
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