<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 FWT{
    public:
    size_t sz;
    vector<int> T;
    FWT(size_t a){
        sz = a;
        T.resize(sz,0);
    }
    int query(int idx){
        int res = 0;
        for(;idx;idx -= (idx&-idx)) res = max(res,T[idx]);
        return res;
    }
    void update(int idx,int val){
        for(;idx<sz;idx+=(idx&-idx)) T[idx] = max(T[idx],val);
    }
};
class Solution {
public:
    int minimumMountainRemovals(vector<int>& nums) {    
        set<int> se; int idx = 0;
        for(auto g:nums) se.insert(g);
        map<int,int> mp;
        for(auto g:se) mp[g]=++idx; // this concludes index compression
        int n = nums.size();
        FWT forward(idx+1),backward(idx+1);
        vector<int> f(n),b(n);
        for(int i=0;i<n;i++){
            f[i] = forward.query(mp[nums[i]]-1) + 1;
            forward.update(mp[nums[i]],f[i]);
        }
        for(int j=n-1;j>=0;j--){
            b[j] = backward.query(mp[nums[j]]-1) + 1;
            backward.update(mp[nums[j]],b[j]);
        }
        int res = nums.size();
        for(int i=0;i<n;i++){
            int deletion = n-f[i]-b[i]+1;
            // cout<<f[i]<<" "<<b[i]<<" "<<deletion<<endl;
            if (f[i]>1 and b[i]>1)
                res=min(res,deletion);
        }
        return res;
    }
};
22600