<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 {
private:
    int dp[1001][1001][2][2];

    int f(int index, int prev, int temp, int num, vector<int> &nums){
        if(index == -1){
            if(temp == 0 && num > 0) return 0;

            return INT_MIN;
        }

        if(dp[index][prev][temp][num] != -1) return dp[index][prev][temp][num];

        if(prev == nums.size()){
            int take = 1 + f(index-1, index, temp, 1, nums);
            int nottake = f(index-1, prev, temp, num, nums);

            return dp[index][prev][temp][num] = max(take, nottake);
        }
        else{
            if(temp == 1){
                int nottake = f(index-1, prev, temp, num, nums);
                int take1 = INT_MIN; int take2 = INT_MIN;

                if(nums[index] > nums[prev]){
                    take1 = 1 + f(index-1, index, 1, 1, nums);
                    take2 = 1 + f(index-1, index, 0, 0, nums);
                }
                
                return dp[index][prev][temp][num] = max(nottake, max(take1, take2));
            }
            else{
                int nottake = f(index-1, prev, temp, num, nums);
                int take = INT_MIN;

                if(nums[index] < nums[prev]){
                    take = 1 + f(index-1, index, 0, 1, nums);
                }

                return dp[index][prev][temp][num] = max(take, nottake);
            }
        }
    }

public:
    int minimumMountainRemovals(vector<int>& nums) {
        // The code is to write the longest Bitonic Subsequence

        // Bitonic subsequence is one which is first increasing than decreasing

        memset(dp, -1, sizeof(dp));

        return nums.size() - f(nums.size()-1, nums.size(), 1, 0, nums);

        return 0;
        // return 0;
    }
};
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