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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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)
          )
  
  

• 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
  }
  

• 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;
  }
  
  

• 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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