1766 Minimum Number Of Removals To Make Mountain Array

Minimum Number of Removals to Make Mountain Array

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.

class Solution:
    def minimumMountainRemovals(self, nums: List[int]) -> int:
        N = len(nums)

        lis_length = [1] * N
        lds_length = [1] * N

        # Stores the length of longest increasing subsequence that ends at i.
        for i in range(N):
            for j in range(i):
                if nums[i] > nums[j]:
                    lis_length[i] = max(lis_length[i], lis_length[j] + 1)

        # Stores the length of longest decreasing subsequence that starts at i.
        for i in range(N - 1, -1, -1):
            for j in range(i + 1, N):
                if nums[i] > nums[j]:
                    lds_length[i] = max(lds_length[i], lds_length[j] + 1)

        min_removals = float("inf")
        for i in range(N):
            if lis_length[i] > 1 and lds_length[i] > 1:
                min_removals = min(
                    min_removals, N - lis_length[i] - lds_length[i] + 1
                )

        return min_removals