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 <= 1001 <= 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).
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(code-box)
