Maximum Alternating Subsequence Sum
<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>
Array; Dynamic Programming
function maxAlternatingSum(nums: number[]): number {
    const n = nums.length;
    const f: number[] = new Array(n + 1).fill(0);
    const g = f.slice();
    for (let i = 1; i <= n; ++i) {
        f[i] = Math.max(g[i - 1] + nums[i - 1], f[i - 1]);
        g[i] = Math.max(f[i - 1] - nums[i - 1], g[i - 1]);
    }
    return Math.max(f[n], g[n]);
}
