<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 {
private:
    vector<int> num;
    const static int MAX = 1000 + 1;
    int memory_lds[MAX];
    int memory_lis[MAX];

    int LIS_END(int idx) {
        if (idx == -1)
            return 0;

        auto& ret = memory_lis[idx];
        if (ret != -1)
            return ret;

        ret = 0; // Maximize among all valid positions after idx
        for (int i = idx - 1; i >= 0; --i) {
            if (num[i] < num[idx])
                ret = max(ret, LIS_END(i)); // Take ith position after idx
        }
        ret += 1; // add idx position
        return ret;
    }
    int LDS_START(int idx) {
        int sz = (int)num.size();
        if (idx == sz)
            return 0;

        auto& ret = memory_lds[idx];
        if (ret != -1)
            return ret;

        ret = 0; // Maximize among all valid positions after idx
        for (int i = idx + 1; i < sz; ++i) {
            if (num[i] < num[idx])
                ret = max(ret, LDS_START(i)); // Take ith position after idx
        }
        ret += 1; // add idx position
        return ret;
    }

public:
    int minimumMountainRemovals(vector<int>& nums) {
        num = nums;
        int sz = nums.size();
        memset(memory_lds, -1, sizeof(memory_lds));
        memset(memory_lis, -1, sizeof(memory_lis));
        int longest_increasing_decreasing = 0;
        for (int i = 0; i < sz; ++i) {
            if (LIS_END(i) == 1 || LDS_START(i) == 1)
                continue;
            int cur = LIS_END(i) + LDS_START(i) - 1;
            longest_increasing_decreasing =
                max(longest_increasing_decreasing, cur);
        }
        return sz - longest_increasing_decreasing;
    }
};
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