How do you solve Count Hills and Valleys in an Array on LeetCode?

Count Hills and Valleys in an Array (LeetCode #2210) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Identifying the closest non-equal neighbors is crucial for determining hills and valleys.

What is the brute force solution for Count Hills and Valleys in an Array?

The brute force approach for Count Hills and Valleys in an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every index to find its closest non-equal neighbors and determines if it is a hill or valley. This is straightforward but inefficient as it involves checking neighbors for each index. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Hills and Valleys in an Array?

Count Hills and Valleys in an Array uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Count Hills and Valleys in an Array?

Clarify the definition of hills and valleys with the interviewer to ensure understanding. Think about edge cases, such as arrays with all equal elements or very small arrays. Consider optimizing your solution after implementing the brute force approach.

What are common mistakes when solving Count Hills and Valleys in an Array?

Failing to check for non-equal neighbors correctly, leading to incorrect counts. Not handling edge cases where the first or last elements do not have both neighbors.

#2210

Count Hills and Valleys in an Array

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the array to identify hills and valleys by keeping track of the last non-equal neighbor. This reduces unnecessary checks and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter for hills and valleys and a variable to track the last non-equal value.
2Step 2: Iterate through the array, checking for hills and valleys only when the current value is different from the last non-equal value.
3Step 3: Update the last non-equal value as you go.
4Step 4: Return the counter.

solution.py18 lines

1def countHillsAndValleys(nums):
2    count = 0
3    n = len(nums)
4    last_non_equal = None
5    for i in range(1, n - 1):
6        if nums[i] != last_non_equal:
7            left, right = i - 1, i + 1
8            while left >= 0 and nums[left] == nums[i]:
9                left -= 1
10            while right < n and nums[right] == nums[i]:
11                right += 1
12            if left >= 0 and right < n:
13                if nums[left] < nums[i] > nums[right]:
14                    count += 1
15                elif nums[left] > nums[i] < nums[right]:
16                    count += 1
17            last_non_equal = nums[i]
18    return count

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, checking each index once. The space complexity is O(1) as we use a constant amount of extra space.

1Identifying the closest non-equal neighbors is crucial for determining hills and valleys.
2Adjacent indices with the same value should be treated as part of the same hill or valley.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.