<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>

class Solution:
    def maxValueAfterReverse(self, nums):
        total = 0
        mn = float('inf')
        mx = float('-inf')

        for i in range(len(nums) - 1):
            a = nums[i]
            b = nums[i + 1]
            total += abs(a - b)
            mn = min(mn, max(a, b))
            mx = max(mx, min(a, b))

        diff = max(0, (mx - mn) * 2)

        for i in range(len(nums) - 1):
            a = nums[i]
            b = nums[i + 1]
            head_diff = -abs(a - b) + abs(nums[0] - b)
            tail_diff = -abs(a - b) + abs(nums[-1] - a)
            diff = max(diff, max(head_diff, tail_diff))

        return total + diff

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