<p>You may recall that an array <code>arr</code> is a <strong>mountain array</strong> if and only if:</p>

<ul>
	<li><code>arr.length &gt;= 3</code></li>
	<li>There exists some index <code>i</code> (<strong>0-indexed</strong>) with <code>0 &lt; i &lt; arr.length - 1</code> such that:
	<ul>
		<li><code>arr[0] &lt; arr[1] &lt; ... &lt; arr[i - 1] &lt; arr[i]</code></li>
		<li><code>arr[i] &gt; arr[i + 1] &gt; ... &gt; arr[arr.length - 1]</code></li>
	</ul>
	</li>
</ul>

<p>Given an integer array <code>nums</code>​​​, return <em>the <strong>minimum</strong> number of elements to remove to make </em><code>nums<em>​​​</em></code><em> </em><em>a <strong>mountain array</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li>It is guaranteed that you can make a mountain array out of <code>nums</code>.</li>
</ul>

class Solution:
    def minimumMountainRemovals(self, nums: List[int]) -> int:
        dp_increase = [0 for _ in nums]
        dp_decrease = [0 for _ in nums]
        self.get_longest_increasing_subseq(nums, dp_increase)
        self.get_longest_decreasing_subseq(nums, dp_decrease)
        res = 0
        for i in range(len(nums)):
            res = max(res, dp_increase[i] + dp_decrease[i] + 1 if dp_increase[i] > 0 and dp_decrease[i] else 0)
        return len(nums) - res

    
    def get_longest_increasing_subseq(self, nums:List[int], dp: List[int]) -> None:
        dp[0] = 0
        for i in range(1, len(nums)):
            for j in range(0, i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)


    def get_longest_decreasing_subseq(self, nums:List[int], dp: List[int]) -> None:
        dp[len(nums) - 1] = 0
        for i in range(len(nums) - 2, -1, -1):
            for j in range(i + 1, len(nums)):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)

16700