LeetCode #2210 — EASY

Count Hills and Valleys in an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,4,1,1,6,5]

Example 2

[6,6,5,5,4,1]

Core Insight

What unlocks the optimal approach

For each index, could you find the closest non-equal neighbors?
Ensure that adjacent indices that are part of the same hill or valley are not double-counted.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2210: Count Hills and Valleys in an Array
class Solution {
    public int countHillValley(int[] nums) {
        int ans = 0, j = 0;
        for (int i = 1; i < nums.length - 1; ++i) {
            if (nums[i] == nums[i + 1]) {
                continue;
            }
            if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
                ++ans;
            }
            if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
                ++ans;
            }
            j = i;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2210: Count Hills and Valleys in an Array
func countHillValley(nums []int) int {
	ans := 0
	for i, j := 1, 0; i < len(nums)-1; i++ {
		if nums[i] == nums[i+1] {
			continue
		}
		if nums[i] > nums[j] && nums[i] > nums[i+1] {
			ans++
		}
		if nums[i] < nums[j] && nums[i] < nums[i+1] {
			ans++
		}
		j = i
	}
	return ans
}

# Accepted solution for LeetCode #2210: Count Hills and Valleys in an Array
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

// Accepted solution for LeetCode #2210: Count Hills and Valleys in an Array
impl Solution {
    pub fn count_hill_valley(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut j = 0;

        for i in 1..nums.len() - 1 {
            if nums[i] == nums[i + 1] {
                continue;
            }
            if nums[i] > nums[j] && nums[i] > nums[i + 1] {
                ans += 1;
            }
            if nums[i] < nums[j] && nums[i] < nums[i + 1] {
                ans += 1;
            }
            j = i;
        }

        ans
    }
}

// Accepted solution for LeetCode #2210: Count Hills and Valleys in an Array
function countHillValley(nums: number[]): number {
    let ans = 0;
    for (let i = 1, j = 0; i < nums.length - 1; ++i) {
        if (nums[i] === nums[i + 1]) {
            continue;
        }
        if (nums[i] > nums[j] && nums[i] > nums[i + 1]) {
            ans++;
        }
        if (nums[i] < nums[j] && nums[i] < nums[i + 1]) {
            ans++;
        }
        j = i;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.