<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:

    void solveOptimal(vector<int> &arr, vector<int> &lis){
        if(arr.size() == 0){
            return;
        }
        vector<int> ans;
        lis.push_back(1);
        ans.push_back(arr[0]);

        for(int i=1; i<arr.size(); i++){
            if(arr[i] > ans.back()){
                //include
                ans.push_back(arr[i]);
                lis.push_back(ans.size());
            }
            else{
                //overwrite
                //find index of just bada element
                int index = lower_bound(ans.begin(), ans.end(), arr[i])-ans.begin();
                ans[index] = arr[i];
                lis.push_back(index+1);
            }
        }
        //return ans.size();
    }

    int minimumMountainRemovals(vector<int>& nums) {
        //longest increasing subsequence, longest decreasing subsequence
        vector<int> lis, lds;
        //make lis and lds array by taking each element as pivot
        solveOptimal(nums, lis);
        reverse(nums.begin(), nums.end());
        solveOptimal(nums, lds);
        reverse(lds.begin(), lds.end());

        //main logic
        int largestMountain = INT_MIN;
        for(int i=0; i<nums.size(); ++i){
            //if(lis[i] || lds[nums.size()-i-1] == 1) continue;
            if(lis[i] > 1 && lds[i] > 1) {
                largestMountain = max(largestMountain, lis[i] + lds[i] - 1);
            }
        }

        int minRemovals = nums.size() - largestMountain;
        return minRemovals;
    }
};
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