Count the Number of Incremovable Subarrays I — LeetCode 2970 Python Solution

The problem

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.

Example

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.

Python solution

class Solution:
    def incremovableSubarrayCount(self, nums: List[int]) -> int:
        i, n = 0, len(nums)
        while i + 1 < n and nums[i] < nums[i + 1]:
            i += 1
        if i == n - 1:
            return n * (n + 1) // 2
        ans = i + 2
        j = n - 1
        while j:
            while i >= 0 and nums[i] >= nums[j]:
                i -= 1
            ans += i + 2
            if nums[j - 1] >= nums[j]:
                break
            j -= 1
        return ans

Complexity

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 2970. Count the Number of Incremovable Subarrays I is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Two Pointers.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

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