Reverse Subarray To Maximize Array Value
<p>You are given an integer array <code>nums</code>. The <em>value</em> of this array is defined as the sum of <code>|nums[i] - nums[i + 1]|</code> for all <code>0 &lt;= i &lt; nums.length - 1</code>.</p>

<p>You are allowed to select any subarray of the given array and reverse it. You can perform this operation <strong>only once</strong>.</p>

<p>Find maximum possible value of the final array.</p>

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

<pre>
<strong>Input:</strong> nums = [2,3,1,5,4]
<strong>Output:</strong> 10
<b>Explanation: </b>By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,9,24,2,1,10]
<strong>Output:</strong> 68
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>The answer is guaranteed to fit in a 32-bit integer.</li>
</ul>
Greedy; Array; Math
class Solution:
    def maxValueAfterReverse(self, nums: List[int]) -> int:
        ans = s = sum(abs(x - y) for x, y in pairwise(nums))
        for x, y in pairwise(nums):
            ans = max(ans, s + abs(nums[0] - y) - abs(x - y))
            ans = max(ans, s + abs(nums[-1] - x) - abs(x - y))
        for k1, k2 in pairwise((1, -1, -1, 1, 1)):
            mx, mi = -inf, inf
            for x, y in pairwise(nums):
                a = k1 * x + k2 * y
                b = abs(x - y)
                mx = max(mx, a - b)
                mi = min(mi, a + b)
            ans = max(ans, s + max(mx - mi, 0))
        return ans
