<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:
    // LIS
    void solveBin(vector<int> &nums, vector<int> &lis)
    {
        vector<int> ans;
        ans.push_back(nums[0]);
        lis.push_back(1);
        for(int i=1; i<nums.size(); i++)
        {
          if(ans.back()<nums[i])
          {
             ans.push_back(nums[i]);
             lis.push_back(ans.size());
          }
          else
          {
            int idx = lower_bound(ans.begin(), ans.end(), nums[i]) - ans.begin();
            ans[idx] = nums[i];
            lis.push_back(idx+1);
          }
        }
    }

    int minimumMountainRemovals(vector<int>& nums) {
        vector<int> lis,lds;
        vector<int> temp=nums;
        reverse(temp.begin(), temp.end());
        solveBin(nums, lis);
        solveBin(temp, lds);
        int ans=INT_MIN;
        reverse(lds.begin(), lds.end());
        for(int i=0; i<lis.size(); i++)
        {
            if(lis[i]==1 || lds[i]==1) continue;
            ans=max(ans, lis[i]+lds[i]-1);
        }
        return nums.size()-ans;
    }
};
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