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, 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

    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