<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:
    vector<int> compute(int N, vector<int> &nums) {
        map<int,int> longest_increasing_seq;
        vector<int> res(N, 1);
        for (int i = 0; i < N; ++i) {
            int best_length = 1;
            for (auto it: longest_increasing_seq) {
                if (it.first >= nums[i]) break;
                best_length = max(best_length, it.second + 1);
            }
            longest_increasing_seq[nums[i]] = best_length;
            res[i] = best_length;
            // cout << i << " " << nums[i] << " " << best_length << "\n";
        }
        return res;
    }

    int minimumMountainRemovals(vector<int>& nums) {
        int N = nums.size();
        vector<int> increasing(N, 1), decreasing(N, 1);
        increasing = compute(N, nums);
        reverse(nums.begin(), nums.end());
        decreasing = compute(N, nums);
        reverse(decreasing.begin(), decreasing.end());
        int ans = N;
        for (int i = 0; i < N; ++i) {
            if (increasing[i] == 1 || decreasing[i] == 1) continue;
            int remove = (i + 1) - increasing[i] + (N - i) - decreasing[i];
            ans = min(ans, remove);
            // cout << i << " "  << nums[i] << " " <<  decreasing[i] << " " << increasing[i] << "\n";
            // cout << i << " r: " << remove << " front: " << (i + 1) - increasing[i] << " back: " << (N - i) - decreasing[i] << "\n";
        }
        return ans;
    }
};
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