1330. Reverse Subarray To Maximize Array Value

You are given an integer array nums. The value of this array is defined as the sum of |nums[i] - nums[i + 1]| for all 0 <= i < nums.length - 1.

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

Find maximum possible value of the final array.

class Solution {
public:
    int maxValueAfterReverse(vector<int>& A) {
        int sum = 0, rev = 0, ma = -1e5, mi = 1e5, n = A.size();
        for(int i = 0; i < n - 1; i++) {
            sum += abs(A[i] - A[i + 1]);
            rev = max(rev, abs(A[0] - A[i + 1]) - abs(A[i] - A[i + 1]));
            rev = max(rev, abs(A[n - 1] - A[i]) - abs(A[i] - A[i + 1]));
            mi = min(mi, max(A[i], A[i + 1]));
            ma = max(ma, min(A[i], A[i + 1]));
        }
        return sum + max(rev, (ma - mi) * 2);
    }
};