README_EN.md

2210. Count Hills and Valleys in an Array

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Description

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

Solutions

Python3

First make the consecutive duplicate value to be unique with the side values.

class Solution:
    def countHillValley(self, nums: List[int]) -> int:
        arr = [nums[0]]
        for v in nums[1:]:
            if v != arr[-1]:
                arr.append(v)
        return sum(
            (arr[i] < arr[i - 1]) == (arr[i] < arr[i + 1])
            for i in range(1, len(arr) - 1)
        )

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

Java

Use two pointers to solve the problem

class Solution {
    public int countHillValley(int[] nums) {
        int ans = 0;
        for (int 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;
    }
}

C++

class Solution {
public:
    int countHillValley(vector<int>& nums) {
        int ans = 0;
        for (int i = 1, j = 0; i < nums.size() - 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;
    }
};

Go

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
}

TypeScript

function countHillValley(nums: number[]): number {
    const n = nums.length;
    let res = 0;
    let prev = nums[0];
    for (let i = 1; i < n - 1; i++) {
        const num = nums[i];
        const next = nums[i + 1];
        if (num == next) {
            continue;
        }
        if ((num > prev && num > next) || (num < prev && num < next)) {
            res += 1;
        }
        prev = num;
    }
    return res;
}

Rust

impl Solution {
    pub fn count_hill_valley(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut res = 0;
        let mut prev = nums[0];
        for i in 1..n - 1 {
            let num = nums[i];
            let next = nums[i + 1];
            if num == next {
                continue;
            }
            if num > prev && num > next || num < prev && num < next {
                res += 1;
            }
            prev = num;
        }
        res
    }
}

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