<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 {
public:
    int maxValueAfterReverse(vector<int>& A) {
        int total = 0, res = 0, min2 = 123456, max2 = -123456, n = A.size();
        for (int i = 0; i < n - 1; ++i) {
            int a = A[i], b = A[i + 1];
            total += abs(a - b);
            res = max(res, abs(A[0] - b) - abs(a - b));
            res = max(res, abs(A[n - 1] - a) - abs(a - b));
            min2 = min(min2, max(a, b));
            max2 = max(max2, min(a, b));
        }
        return total + max(res, (max2 - min2) * 2);
    }
};
43000