Description
You are given a 0-indexed array of positive integers nums.
A subarray of nums is called incremovable if nums becomes strictly increasing on removing the subarray. For example, the subarray [3, 4] is an incremovable subarray of [5, 3, 4, 6, 7] because removing this subarray changes the array [5, 3, 4, 6, 7] to [5, 6, 7] which is strictly increasing.
Return the total number of incremovable subarrays of nums.
Note that an empty array is considered strictly increasing.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [1,2,3,4] Output: 10 Explanation: The 10 incremovable subarrays are: [1], [2], [3], [4], [1,2], [2,3], [3,4], [1,2,3], [2,3,4], and [1,2,3,4], because on removing any one of these subarrays nums becomes strictly increasing. Note that you cannot select an empty subarray.
Example 2:
Input: nums = [6,5,7,8] Output: 7 Explanation: The 7 incremovable subarrays are: [5], [6], [5,7], [6,5], [5,7,8], [6,5,7] and [6,5,7,8]. It can be shown that there are only 7 incremovable subarrays in nums.
Example 3:
Input: nums = [8,7,6,6] Output: 3 Explanation: The 3 incremovable subarrays are: [8,7,6], [7,6,6], and [8,7,6,6]. Note that [8,7] is not an incremovable subarray because after removing [8,7] nums becomes [6,6], which is sorted in ascending order but not strictly increasing.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109
Solution
Python3
class Solution:
def incremovableSubarrayCount(self, nums: List[int]) -> int:
N = len(nums)
right = [nums[-1]]
for i in range(N - 2, -1, -1):
if nums[i] < nums[i + 1]:
right.append(nums[i])
else:
break
right.reverse()
if len(right) == N: return N * (N + 1) // 2
res = len(right) + 1
last = -1
for i in range(0, N - len(right)):
if nums[i] <= last: break
last = nums[i]
res += len(right) - bisect_right(right, nums[i]) + 1
return res