<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();
        vector<int> dp1(n, 1), dp2(n, 1);

        // for lis from front
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i] && dp1[j] + 1 > dp1[i]) {
                    dp1[i] = dp1[j] + 1;
                }
            }
        }

        // for lis from back
        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j > i; j--) {
                if (nums[j] < nums[i] && dp2[j] + 1 > dp2[i]) {
                    dp2[i] = dp2[j] + 1;
                }
            }
        }

        // biotonic
        int maxi = 0;
        for (int i = 0; i < n; i++) {
            if (dp1[i] == 1 || dp2[i] == 1)
                continue;
            maxi = max(maxi, dp1[i] + dp2[i] - 1);
        }
        return n - maxi;
    }
};
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