You may recall that an array arr is a mountain array if and only if:
arr.length >= 3
There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given an integer array nums, return the minimum number of elements to remove to make numsa mountain array.
Example 1:
Input: nums = [1,3,1]
Output: 0
Explanation: The array itself is a mountain array so we do not need to remove any elements.
Example 2:
Input: nums = [2,1,1,5,6,2,3,1]
Output: 3
Explanation: One solution is to remove the elements at indices 0, 1, and 5, making the array nums = [1,5,6,3,1].
Constraints:
3 <= nums.length <= 1000
1 <= nums[i] <= 109
It is guaranteed that you can make a mountain array out of nums.
Solutions
Solution 1: Dynamic Programming
This problem can be transformed into finding the longest increasing subsequence and the longest decreasing subsequence.
We define $left[i]$ as the length of the longest increasing subsequence ending with $nums[i]$, and define $right[i]$ as the length of the longest decreasing subsequence starting with $nums[i]$.
Then the final answer is $n - \max(left[i] + right[i] - 1)$, where $1 \leq i \leq n$, and $left[i] \gt 1$ and $right[i] \gt 1$.
The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.