Introduction

In this blog post, we will explore how to solve the LeetCode problem "2970" 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 <= 50 1 <= nums[i] <= 50

Explanation

Here's a solution to the problem:

• High-level approach:

  • Iterate through all possible subarrays of the given array nums.
  • For each subarray, create a new array by removing that subarray from the original array.
  • Check if the resulting array is strictly increasing. If it is, increment the count of incremovable subarrays.

• Complexity:

  • Runtime: O(n^3), where n is the length of the input array.
  • Storage: O(n)

Code

    def incremovableSubarrayCount(nums):
    """
    Calculates the number of incremovable subarrays in a given array.

    Args:
        nums: A list of positive integers.

    Returns:
        The total number of incremovable subarrays of nums.
    """

    n = len(nums)
    count = 0

    for i in range(n):
        for j in range(i, n):
            # Create a new array by removing the subarray nums[i:j+1]
            new_nums = nums[:i] + nums[j+1:]

            # Check if the new array is strictly increasing
            is_increasing = True
            if len(new_nums) > 1:
                for k in range(len(new_nums) - 1):
                    if new_nums[k] >= new_nums[k+1]:
                        is_increasing = False
                        break

            if is_increasing:
                count += 1

    return count