<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 n = nums.size();
        if (n < 3) return 0;  // Not enough elements to form a mountain
        
        vector<int> dp1(n, 1), dp2(n, 1);
        set<int> lis;
        
        // Compute LIS length from left to right
        for (int i = 0; i < n; ++i) {
            int x = nums[i];
            auto it = lis.upper_bound(x);
            if (it != lis.end()) {
                lis.erase(it);
            }
            lis.insert(x);
            dp1[i] = lis.size();
        }
        
        lis.clear();
        // Compute LIS length from right to left
        for (int i = n - 1; i >= 0; --i) {
            int x = nums[i];
            auto it = lis.upper_bound(x);
            if (it != lis.end()) {
                lis.erase(it);
            }
            lis.insert(x);
            dp2[i] = lis.size();
        }
        
        int res = n;
        // Find the minimum number of removals
        for (int i = 1; i < n - 1; ++i) {
            if (dp1[i] > 1 && dp2[i] > 1) {
                res = min(res, (n - (dp1[i] + dp2[i]) + 1));
            }
        }
        
        return res;
    }
};

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