Given an integer array nums, count how many hills and valleys it contains, and return that count.

For an index i, let the closest non equal element to the left of i be the first index l < i such that nums[l] != nums[i] and for every k with l < k < i, nums[k] == nums[i]. Similarly, let the closest non equal element to the right of i be the first index r > i such that nums[r] != nums[i] and for every k with i < k < r, nums[k] == nums[i].

Index i is a hill if both such neighbors exist and nums[i] is strictly greater than both of them. Index i is a valley if both such neighbors exist and nums[i] is strictly smaller than both of them.

If multiple consecutive indices have the same value and they form a single hill or a single valley with respect to their closest non equal neighbors, count that hill or valley only once.

Implement int countHillValley(self, nums) to return the total number of hills and valleys in nums.

Note: When determining hills and valleys, equal adjacent values are ignored by looking outward to the nearest different values.

Constraints:

• $3 \leq$ nums.length $\leq 100$

• $1 \leq$ nums[i] $\leq 100$

Input: nums = [1, 3, 2]-List[int]
Output: 1-int
Explanation: Index 1 (value 3) has nearest different neighbors 1 and 2, and since 3 is greater than both, it forms one hill.

Input: nums = [5, 5, 5, 5, 5]-List[int]
Output: 0-int
Explanation: All values are equal, so no index has two closest non-equal neighbors, resulting in no hills or valleys.

Input: nums = [3, 3, 1, 2, 2, 4, 4, 2]-List[int]
Output: 2-int
Explanation: Ignoring equal-adjacent runs, the sequence forms one valley at the 1 (between 3 and 2) and one hill at the 4 (between 2 and 2), counted once per plateau.

Input: nums = [2, 2, 4, 4, 1, 1, 3, 3, 1]-List[int]
Output: 3-int
Explanation: Collapsing equal runs gives 2, 4, 1, 3, 1 where 4 is a hill, 1 (between 4 and 3) is a valley, and 3 is a hill, totaling three.

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