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] > ... > arr[arr.length - 1]`

Given an integer array `nums`​​​, return _the **minimum** number of elements to remove to make_ `nums_​​​_` _a **mountain array**._

**Example 1:**

**Input:** nums = \[1,3,1\]
**Output:** 0
**Explanation:** The array itself is a mountain array so we do not need to remove any elements.

**Example 2:**

**Input:** nums = \[2,1,1,5,6,2,3,1\]
**Output:** 3
**Explanation:** One solution is to remove the elements at indices 0, 1, and 5, making the array nums = \[1,5,6,3,1\].

**Constraints:**

*   `3 <= nums.length <= 1000`
*   `1 <= nums[i] <= 109`
*   It is guaranteed that you can make a mountain array out of `nums`.


def minimumMountainRemovals(nums):
    n = len(nums)
    left, right = [0] * n, [0] * n
    for i in range(1, n):
        left[i] = left[i - 1] + int(nums[i] > nums[i - 1])
        right[n - i - 1] = right[n - i] + int(nums[n - i - 1] > nums[n - i])
    ans = n - 3
    for i in range(1, n - 1):
        ans = min(ans, n - 1 - left[i] - right[i])
    return ans
