<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>& nums) {
        int n = nums.size();
        int total = 0;
        for (int i = 0; i < n - 1; ++i) {
            total += abs(nums[i] - nums[i + 1]);
        }

        int gain = 0;
        for (int i = 1; i < n - 1; ++i) {
            gain = max(gain, abs(nums[0] - nums[i + 1]) - abs(nums[i] - nums[i + 1]));
            gain = max(gain, abs(nums[n-1] - nums[i - 1]) - abs(nums[i] - nums[i - 1]));
        }

        int min1 = INT_MAX, max1 = INT_MIN;
        for (int i = 0; i < n - 1; ++i) {
            int x = nums[i], y = nums[i + 1];
            min1 = min(min1, max(x, y));
            max1 = max(max1, min(x, y));
        }

        gain = max(gain, 2 * (max1 - min1));
        return total + gain;
    }
};
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