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2210. Count Hills and Valleys in an Array

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

Solution 1: Traversal

We initialize a pointer $j$ to point to the position with index $0$, and then traverse the array in the range $[1, n-1]$. For each position $i$:

  • If $nums[i] = nums[i+1]$, then skip.
  • Otherwise, if $nums[i]$ is greater than $nums[j]$ and $nums[i]$ is greater than $nums[i+1]$, then $i$ is a peak; if $nums[i]$ is less than $nums[j]$ and $nums[i]$ is less than $nums[i+1]$, then $i$ is a valley.
  • Then, we update $j$ to $i$ and continue to traverse.

After the traversal, we can get the number of peaks and valleys.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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