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Count Hills and Valleys in an Array

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

Solution Explanation for LeetCode 2210: Count Hills and Valleys in an Array

This problem asks us to count the number of "hills" and "valleys" in a given integer array. A hill is a point where its immediate neighbors (left and right) are strictly smaller, and a valley is a point where its immediate neighbors are strictly larger. Crucially, consecutive identical values are considered part of the same hill or valley.

Approach

The most efficient approach is a single pass through the array. We don't need to store any extra information beyond a pointer to the previous non-equal element.

  1. Initialization: We start with a counter ans initialized to 0 (for hills and valleys) and a pointer j to 0 (index of the previous non-equal element).

  2. Iteration: We iterate through the array from index 1 to n-2 (where n is the array length). We skip consecutive equal elements.

  3. Hill/Valley Check: For each index i, we check if it forms a hill or a valley. This involves comparing nums[i] with both nums[j] (the previous non-equal element) and nums[i+1].

    • Hill: If nums[i] > nums[j] and nums[i] > nums[i+1], we increment ans.
    • Valley: If nums[i] < nums[j] and nums[i] < nums[i+1], we increment ans.

  4. Update j: After checking, we update j to i, so it points to the current non-equal element for the next iteration.

Time and Space Complexity

  • Time Complexity: O(n), where n is the length of the input array. We perform a single pass through the array.
  • Space Complexity: O(1). We use only a constant amount of extra space to store variables (ans and j).

Code Implementation (Python)

class Solution:
    def countHillValley(self, nums: List[int]) -> int:
        ans = j = 0
        for i in range(1, len(nums) - 1):
            if nums[i] == nums[i + 1]:
                continue
            if nums[i] > nums[j] and nums[i] > nums[i + 1]:
                ans += 1
            if nums[i] < nums[j] and nums[i] < nums[i + 1]:
                ans += 1
            j = i
        return ans

This Python code directly implements the algorithm described above. The other languages (Java, C++, Go, TypeScript, Rust) would follow a very similar structure, adapting syntax and data types accordingly. Refer to the original response for the code in these languages.