Minimum Number of Removals to Make Mountain Array — LeetCode 1671 Python Solution

The problem

You may recall that an array arr is a mountain array if and only if: arr.length >= 3 There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that: arr[0] < arr[1] < ... < arr[i - 1] < arr[i] arr[i] > arr[i + 1] > ...

Example

Input

nums = [1,3,1]

Output

0

Explanation

The array itself is a mountain array so we do not need to remove any elements.

Python solution

class Solution:
    def minimumMountainRemovals(self, nums: List[int]) -> int:
        n = len(nums)
        left = [1] * n
        right = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[i] > nums[j]:
                    left[i] = max(left[i], left[j] + 1)
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                if nums[i] > nums[j]:
                    right[i] = max(right[i], right[j] + 1)
        return n - max(a + b - 1 for a, b in zip(left, right) if a > 1 and b > 1)

Complexity

Pattern: Monotonic Stack

Answer "what is the next greater element" for every position in one pass. LeetCode 1671. Minimum Number of Removals to Make Mountain Array is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.

The monotonic stack guide has the Python template for the pattern and the 225 LeetCode problems that use it.

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