Minimum Number of Removals to Make Mountain Array
<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>
Greedy; Array; Binary Search; Dynamic Programming
impl Solution {
    pub fn minimum_mountain_removals(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut left = vec![1; n];
        let mut right = vec![1; n];
        for i in 1..n {
            for j in 0..i {
                if nums[i] > nums[j] {
                    left[i] = left[i].max(left[j] + 1);
                }
            }
        }
        for i in (0..n - 1).rev() {
            for j in i + 1..n {
                if nums[i] > nums[j] {
                    right[i] = right[i].max(right[j] + 1);
                }
            }
        }

        let mut ans = 0;
        for i in 0..n {
            if left[i] > 1 && right[i] > 1 {
                ans = ans.max(left[i] + right[i] - 1);
            }
        }

        (n as i32) - ans
    }
}
