<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> leftToRight, rightToLeft;
    unordered_map<int, int> lenLeftToRight, lenRightToLeft;
    int n;

    int minimumMountainRemovals(vector<int>& nums) {
        n = nums.size();

        // LIS left to right: O(n * log n)
        for (int i = 0; i < n; i++) {
            if (leftToRight.empty() || nums[i] > leftToRight.back()) {
                leftToRight.push_back(nums[i]);
                lenLeftToRight[i] = leftToRight.size();
            }
            else {
                int idx = lower_bound(leftToRight.begin(), leftToRight.end(), nums[i]) - leftToRight.begin();
                leftToRight[idx] = nums[i];
                lenLeftToRight[i] = idx + 1;
            }
        }

        // LIS right to left: O(n * log n)
        for (int i = n - 1; i >= 0; i--) {
            if (rightToLeft.empty() || nums[i] > rightToLeft.back()) {
                rightToLeft.push_back(nums[i]);
                lenRightToLeft[i] = rightToLeft.size();
            }
            else {
                int idx = lower_bound(rightToLeft.begin(), rightToLeft.end(), nums[i]) - rightToLeft.begin();
                rightToLeft[idx] = nums[i];
                lenRightToLeft[i] = idx + 1;
            }
        }

        // Collate longest bitonic subsequence length
        int longestBitonicSubseqLen = 0;
        for (int i = 0; i < n; i++) {
            if (lenLeftToRight[i] > 1 && lenRightToLeft[i] > 1)
                longestBitonicSubseqLen = max(longestBitonicSubseqLen, 
                    lenLeftToRight[i] + lenRightToLeft[i] - 1);
        }
        
        // Compute ans as total len - longestBitonicSubseqLen
        return n - longestBitonicSubseqLen;
    }
};
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