1671. Minimum Number of Removals to Make Mountain Array

Problem Link

Description

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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.

Solution

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Python3

class Solution:
    def minimumMountainRemovals(self, nums: List[int]) -> int:
        N = len(nums)
 
        lis = [1] * N
        for i in range(N):
            for j in range(i):
                if nums[i] > nums[j]:
                    lis[i] = max(lis[i], lis[j] + 1)
        
        lds = [1] * N
        for i in range(N - 1, -1, -1):
            for j in range(i + 1, N):
                if nums[i] > nums[j]:
                    lds[i] = max(lds[i], lds[j] + 1)
 
        maxLength = 0
        for i in range(1, N - 1):
            if lis[i] > 1 and lds[i] > 1:
                maxLength = max(maxLength, lis[i] + lds[i] - 1)
 
        return N - maxLength