<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:
    vector<int> lisLeft(vector<int>& nums,int n){
        vector<int>pre;
        vector<int>ans(n,0);
        pre.push_back(nums[0]);
        
        for(int i=1;i<n;i++){
            if(nums[i]>pre.back()){
                {
                    pre.push_back(nums[i]);
                    ans[i]=pre.size();
                }
            }else {
                int lb = lower_bound(pre.begin(), pre.end(), nums[i]) - pre.begin();
                pre[lb] = nums[i];
                ans[i] = lb + 1; // update the length of LIS up to this point
            }
        }
        return ans;
    }
    vector<int> lisright(vector<int>& nums,int n){
        vector<int>pre;
        pre.push_back(nums[0]);
        vector<int>ans(n,0);
        for(int i=1;i<n;i++){
            if(nums[i]>pre.back()){
                {
                    pre.push_back(nums[i]);
                    ans[i]=pre.size();
                }
            }else {
                int lb = lower_bound(pre.begin(), pre.end(), nums[i]) - pre.begin();
                pre[lb] = nums[i];
                ans[i] = lb + 1; // update the length of LIS up to this point
            }
        }
        return ans;
    }
    int minimumMountainRemovals(vector<int>& nums) {
        int n=nums.size();
        vector<int> pre=lisLeft(nums,n);
        vector<int>v=nums;
        reverse(v.begin(),v.end());
        vector<int> suf=lisright(v,n);
        reverse(suf.begin(), suf.end());
        int ans=0;
        for(int i=0;i<n;i++){
            if(pre[i]>1 && suf[i]>1){
                ans=max(ans,pre[i]+suf[i]-1);
            }
        }
        return n-ans;
    }
};
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