【leetcode】1671. Minimum Number of Removals to Make Mountain Array

题目如下：

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 nums​​​ a 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.

解题思路：动态规划。先假设dp_left[i] = v 表示第i个元素为最高点的时候，左半部分需要删除的元素数量的最小值。如果前一个保留的元素下标为j，那么表示需要删除掉(i-j+1)个元素，同时可以得到状态转移方程： dp[i] = max(dp[j] + (i-j+1)) (j <= i)。右半部分也同理，最需要求出左半部分和右半部分和最小的i即可。

代码如下：

class Solution(object):
    def minimumMountainRemovals(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        dp_left = range(len(nums))
        for i in range(1,len(dp_left)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp_left[i] = min(dp_left[i],dp_left[j] + (i-j-1))
        
        dp_right = range(len(nums)-1,-1,-1)
        for i in range(len(dp_right)-2,-1,-1):
            for j in range(len(dp_right)-1,i,-1):
                if nums[i] > nums[j]:
                    dp_right[i] = min(dp_right[i], dp_right[j] + (j - i - 1))

        res = len(nums) - 3
        for i in range(1,len(dp_left)-1):
            if dp_left[i] != i and dp_right[i] != (len(dp_right) - i - 1):
                res = min(res,dp_left[i] + dp_right[i])
        return res