<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 {
public:
    int minimumMountainRemovals(vector<int>& nums) {
        int last=-1;
        int cnt=0;
        int id=0;
        int maxi=0;
        int maxi1=0;
        vector<vector<int>> dp(nums.size(), vector<int>(3, INT_MAX));
        for(int i=0; i<nums.size(); i++)
        {
            dp[i][0]=min(dp[i][0], i);
            // if(i>0 && dp[i][1]!=INT_MAX) dp[i][2]=min(dp[i][2], dp[i][1]);
            for(int j=i+1; j<nums.size(); j++)
            {
                if(nums[j]>nums[i])
                {
                    dp[j][0]=min(dp[j][0], dp[i][0]+(j-i-1));
                    dp[j][1]=min(dp[j][1], dp[i][0]+(j-i-1));
                }
                if(nums[j]<nums[i] && dp[i][1]!=INT_MAX)
                {
                    dp[j][1]=min(dp[j][1], dp[i][1]+(j-i-1));
                    dp[j][2]=min(dp[j][2], dp[i][1]+(j-i-1));
                }
                if(dp[i][2]!=INT_MAX)
                {
                    dp[j][2]=min(dp[j][2], dp[i][2]+(j-i));
                }
            }
        }
        return dp[dp.size()-1][2];
    }
};
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