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Count the Number of Incremovable Subarrays I

Difficulty: Easy

Problem 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. Return the total number of incremovable subarrays of nums.

Key Insights

  • A subarray is considered incremovable if its removal results in a strictly increasing array.
  • The problem can be solved by finding subarrays that do not break the increasing order of the surrounding elements.
  • Every single element is an incremovable subarray by itself.
  • The maximum length of a subarray that can be removed depends on the surrounding elements' values.

Space and Time Complexity

Time Complexity: O(n^2)
Space Complexity: O(1)

Solution

To solve the problem, we can utilize a brute force approach which checks every possible subarray and determines if removing it results in a strictly increasing array.

  1. Iterate through each possible subarray using two nested loops.
  2. For each subarray defined by indices i and j, check if the removal of that subarray results in a strictly increasing order for the elements before index i and after index j.
  3. Count all valid subarrays that satisfy this condition.

This approach leverages a simple enumeration technique to identify valid subarrays.

Code Solutions

def count_incremovable_subarrays(nums):
    n = len(nums)
    count = 0

    for i in range(n):
        for j in range(i, n):
            # Check if removing nums[i:j+1] leads to a strictly increasing array
            left_valid = (i == 0 or nums[i-1] < nums[j])  # Check if the left side is valid
            right_valid = (j == n-1 or nums[i] < nums[j+1])  # Check if the right side is valid
            
            if left_valid and right_valid:
                count += 1

    return count
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