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Commit 19fa5d6

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Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.7 MB (64.44%)
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. An index <code>i</code> is part of a <strong>hill</strong> in <code>nums</code> if the closest non-equal neighbors of <code>i</code> are smaller than <code>nums[i]</code>. Similarly, an index <code>i</code> is part of a <strong>valley</strong> in <code>nums</code> if the closest non-equal neighbors of <code>i</code> are larger than <code>nums[i]</code>. Adjacent indices <code>i</code> and <code>j</code> are part of the <strong>same</strong> hill or valley if <code>nums[i] == nums[j]</code>.</p>
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<p>Note that for an index to be part of a hill or valley, it must have a non-equal neighbor on <strong>both</strong> the left and right of the index.</p>
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<p>Return <i>the number of hills and valleys in </i><code>nums</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,4,1,1,6,5]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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At index 0: There is no non-equal neighbor of 2 on the left, so index 0 is neither a hill nor a valley.
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At index 1: The closest non-equal neighbors of 4 are 2 and 1. Since 4 &gt; 2 and 4 &gt; 1, index 1 is a hill.
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At index 2: The closest non-equal neighbors of 1 are 4 and 6. Since 1 &lt; 4 and 1 &lt; 6, index 2 is a valley.
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At index 3: The closest non-equal neighbors of 1 are 4 and 6. Since 1 &lt; 4 and 1 &lt; 6, index 3 is a valley, but note that it is part of the same valley as index 2.
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At index 4: The closest non-equal neighbors of 6 are 1 and 5. Since 6 &gt; 1 and 6 &gt; 5, index 4 is a hill.
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At index 5: There is no non-equal neighbor of 5 on the right, so index 5 is neither a hill nor a valley.
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There are 3 hills and valleys so we return 3.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [6,6,5,5,4,1]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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At index 0: There is no non-equal neighbor of 6 on the left, so index 0 is neither a hill nor a valley.
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At index 1: There is no non-equal neighbor of 6 on the left, so index 1 is neither a hill nor a valley.
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At index 2: The closest non-equal neighbors of 5 are 6 and 4. Since 5 &lt; 6 and 5 &gt; 4, index 2 is neither a hill nor a valley.
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At index 3: The closest non-equal neighbors of 5 are 6 and 4. Since 5 &lt; 6 and 5 &gt; 4, index 3 is neither a hill nor a valley.
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At index 4: The closest non-equal neighbors of 4 are 5 and 1. Since 4 &lt; 5 and 4 &gt; 1, index 4 is neither a hill nor a valley.
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At index 5: There is no non-equal neighbor of 1 on the right, so index 5 is neither a hill nor a valley.
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There are 0 hills and valleys so we return 0.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 &lt;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
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</ul>
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# Time: O(n)
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# Space: O(n)
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class Solution:
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def countHillValley(self, nums: List[int]) -> int:
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# Remove consecutive duplicates
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filtered = []
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for num in nums:
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if not filtered or filtered[-1] != num:
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filtered.append(num)
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count = 0
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for i in range(1, len(filtered) - 1):
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# Hill condition
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if filtered[i] > filtered[i - 1] and filtered[i] > filtered[i + 1]:
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count += 1
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# Valley condition
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elif filtered[i] < filtered[i - 1] and filtered[i] < filtered[i + 1]:
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count += 1
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return count
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