# Solving Leetcode Interviews in Seconds with AI: Count Hills and Valleys in an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2210" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a 0-indexed integer array nums. An index i is part of a hill in nums if the closest non-equal neighbors of i are smaller than nums[i]. Similarly, an index i is part of a valley in nums if the closest non-equal neighbors of i are larger than nums[i]. Adjacent indices i and j are part of the same hill or valley if nums[i] == nums[j]. Note that for an index to be part of a hill or valley, it must have a non-equal neighbor on both the left and right of the index. Return the number of hills and valleys in nums.   Example 1:  Input: nums = [2,4,1,1,6,5] Output: 3 Explanation: At index 0: There is no non-equal neighbor of 2 on the left, so index 0 is neither a hill nor a valley. At index 1: The closest non-equal neighbors of 4 are 2 and 1. Since 4 > 2 and 4 > 1, index 1 is a hill.  At index 2: The closest non-equal neighbors of 1 are 4 and 6. Since 1 < 4 and 1 < 6, index 2 is a valley. At index 3: The closest non-equal neighbors of 1 are 4 and 6. Since 1 < 4 and 1 < 6, index 3 is a valley, but note that it is part of the same valley as index 2. At index 4: The closest non-equal neighbors of 6 are 1 and 5. Since 6 > 1 and 6 > 5, index 4 is a hill. At index 5: There is no non-equal neighbor of 5 on the right, so index 5 is neither a hill nor a valley.  There are 3 hills and valleys so we return 3.  Example 2:  Input: nums = [6,6,5,5,4,1] Output: 0 Explanation: At index 0: There is no non-equal neighbor of 6 on the left, so index 0 is neither a hill nor a valley. At index 1: There is no non-equal neighbor of 6 on the left, so index 1 is neither a hill nor a valley. At index 2: The closest non-equal neighbors of 5 are 6 and 4. Since 5 < 6 and 5 > 4, index 2 is neither a hill nor a valley. At index 3: The closest non-equal neighbors of 5 are 6 and 4. Since 5 < 6 and 5 > 4, index 3 is neither a hill nor a valley. At index 4: The closest non-equal neighbors of 4 are 5 and 1. Since 4 < 5 and 4 > 1, index 4 is neither a hill nor a valley. At index 5: There is no non-equal neighbor of 1 on the right, so index 5 is neither a hill nor a valley. There are 0 hills and valleys so we return 0.    Constraints:  3 <= nums.length <= 100 1 <= nums[i] <= 100  

	# Explanation
	Here's the breakdown of the solution:

*   **Remove Consecutive Duplicates:** The core idea is to first reduce the input array by removing consecutive duplicate numbers. This simplifies the hill/valley detection as we only need to compare with immediate neighbors.

*   **Iterate and Compare:** Once we have the simplified array, we iterate from the second element to the second-to-last element. For each element, we check if it's a hill (greater than both neighbors) or a valley (smaller than both neighbors).

*   **Count Hills and Valleys:** We maintain a counter to keep track of the number of hills and valleys found.

*   **Runtime and Storage Complexity:**
    *   Runtime: O(n), where n is the length of the input array.
    *   Storage: O(n) in the worst case, when the array `nums` has no consecutive duplicates.

	
	# Code
	```python
	    def countHillValley(nums):
        """
        Counts the number of hills and valleys in an array.

        Args:
            nums: A list of integers.

        Returns:
            The number of hills and valleys in nums.
        """
        filtered_nums = []
        for num in nums:
            if not filtered_nums or num != filtered_nums[-1]:
                filtered_nums.append(num)

        count = 0
        for i in range(1, len(filtered_nums) - 1):
            if (filtered_nums[i] > filtered_nums[i - 1] and filtered_nums[i] > filtered_nums[i + 1]) or \
               (filtered_nums[i] < filtered_nums[i - 1] and filtered_nums[i] < filtered_nums[i + 1]):
                count += 1

        return count
	```