Leetcode 1671. Minimum Number of Removals to Make Mountain Array

迷南。 2022-09-09 00:13 42阅读 0赞

文章作者：Tyan  
博客：[noahsnail.com][] | [CSDN][] | [简书][Link 1]

## 1. Description ##

![Minimum Number of Removals to Make Mountain Array][]

## 2. Solution ##

\*\*解析：\*\*Version 1，分别以数组中的元素作为中心点，在左右两侧分别求最长递增子序列，根据左右两侧的最长递增子序列的长度求出山脉的长度，则要删除的元素个数为数组长度减去最长的山脉长度，速度太慢。Version 2在Version 1的基础上进行了优化，分别求出数组正序和逆序各个位置的最长递增子序列，然后跟Version 1类似，累加左右对应位置的最长递增子序列的长度，即为山脉的长度，则要删除的元素个数为数组长度减去最长的山脉长度，速度明显有了大幅提升。

*  Version 1

class Solution:
        def minimumMountainRemovals(self, nums: List[int]) -> int:
            n = len(nums)
            maximum = 0
            for index in range(1, n - 1):
                left = [nums[index]]
                for i in range(index - 1, -1, -1):
                    if nums[i] >= nums[index] or nums[i] == left[0]:
                        continue
                    elif nums[i] < left[0]:
                        bisect.insort(left, nums[i])
                    else:
                        pos = bisect.bisect(left, nums[i])
                        left[pos-1] = nums[i]
                if len(left) < 2:
                    continue
                right = [nums[index]]
                for i in range(index+1, n):
                    if nums[i] >= nums[index] or nums[i] == right[0]:
                        continue
                    if nums[i] < right[0]:
                        bisect.insort(right, nums[i])
                    else:
                        pos = bisect.bisect(right, nums[i])
                        right[pos-1] = nums[i]
                if len(right) > 1:
                    maximum = max(maximum, len(left) + len(right) - 1)
            return n - maximum

*  Version 2

class Solution:
        def minimumMountainRemovals(self, nums: List[int]) -> int:
            n = len(nums)
            maximum = 0
            left = self.LIS(nums)
            right = self.LIS(nums[::-1])
            for i in range(n):
                if left[i] > 1 and right[n-i-1] > 1:
                    maximum = max(maximum, left[i] + right[n-i-1] - 1)
            return n - maximum
    
    
        def LIS(self, nums):
            n = len(nums)
            dp = [1] * n
            arr = [nums[0]]
            for i in range(1, n):
                if nums[i] > arr[-1]:
                    arr.append(nums[i])
                else:
                    pos = bisect.bisect_left(arr, nums[i])
                    arr[pos] = nums[i]
                dp[i] = len(arr)
            return dp

## Reference ##

1.  [https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/][https_leetcode.com_problems_minimum-number-of-removals-to-make-mountain-array]

[noahsnail.com]: http://noahsnail.com
[CSDN]: http://blog.csdn.net/quincuntial
[Link 1]: http://www.jianshu.com/users/7731e83f3a4e/latest_articles
[Minimum Number of Removals to Make Mountain Array]: /images/20220829/7b54edc45c1d427bb9c879bf14827292.png
[https_leetcode.com_problems_minimum-number-of-removals-to-make-mountain-array]: https://leetcode.com/problems/minimum-number-of-removals-to-make-mountain-array/