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 {
public:
    int maxValueAfterReverse(vector<int>& nums) {
        int n = nums.size();
        int s = 0;
        for (int i = 0; i < n - 1; ++i) {
            s += abs(nums[i] - nums[i + 1]);
        }
        int ans = s;
        for (int i = 0; i < n - 1; ++i) {
            ans = max(ans, s + abs(nums[0] - nums[i + 1]) - abs(nums[i] - nums[i + 1]));
            ans = max(ans, s + abs(nums[n - 1] - nums[i]) - abs(nums[i] - nums[i + 1]));
        }
        int dirs[5] = {1, -1, -1, 1, 1};
        const int inf = 1 << 30;
        for (int k = 0; k < 4; ++k) {
            int k1 = dirs[k], k2 = dirs[k + 1];
            int mx = -inf, mi = inf;
            for (int i = 0; i < n - 1; ++i) {
                int a = k1 * nums[i] + k2 * nums[i + 1];
                int b = abs(nums[i] - nums[i + 1]);
                mx = max(mx, a - b);
                mi = min(mi, a + b);
            }
            ans = max(ans, s + max(0, mx - mi));
        }
        return ans;
    }
};