# Solving Leetcode Interviews in Seconds with AI: Count the Number of Incremovable Subarrays II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2972" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 <= 105 1 <= nums[i] <= 109  

	# Explanation
	Here's the breakdown of the solution:

*   **Two Pointers:** Use two pointers, `left` and `right`, to define the subarray to be removed. The `left` pointer starts from the beginning of the array, and the `right` pointer starts from the end.
*   **Check Incremovability:** For each possible subarray defined by `left` and `right`, check if removing it results in a strictly increasing array. This involves combining the elements before `left` and after `right` and verifying the strictly increasing property.
*   **Count Incremovable Subarrays:** Increment the count for each incremovable subarray found.

*   **Runtime Complexity:** O(n^2)
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def incremovableSubarrayCount(nums):
    n = len(nums)
    count = 0

    for i in range(n):
        for j in range(i, n):
            temp_nums = []
            for k in range(n):
                if k < i or k > j:
                    temp_nums.append(nums[k])

            is_increasing = True
            if len(temp_nums) > 1:
                for k in range(len(temp_nums) - 1):
                    if temp_nums[k] >= temp_nums[k + 1]:
                        is_increasing = False
                        break
            
            if is_increasing:
                count += 1

    return count
	```