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Count the Number of Incremovable Subarrays II

Difficulty: Hard

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 removing it results in a strictly increasing array.
  • To find incremovable subarrays, we need to identify segments of the array where the removal does not violate the strictly increasing property.
  • The problem can be approached using two pointers or binary search to efficiently find the boundaries of potential subarrays.

Space and Time Complexity

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

Solution

To solve the problem, we can utilize a two-pointer technique to identify the segments of the array that can be removed while maintaining the strictly increasing order of the remaining elements. We will iterate through the array to track the lengths of increasing and decreasing segments, counting the valid subarrays as we go.

  1. Traverse the array to find the lengths of segments that can be removed.
  2. For each segment, calculate the number of valid incremovable subarrays based on the length of the segment.
  3. Accumulate the counts of valid subarrays from all segments.

Code Solutions

def count_incremovable_subarrays(nums):
    n = len(nums)
    total_count = 0
    
    # Step 1: Count incremovable subarrays
    i = 0
    while i < n:
        # Start of a new segment
        j = i
        while j < n - 1 and nums[j] < nums[j + 1]:
            j += 1
            
        # Count subarrays in this increasing segment
        length = j - i + 1
        total_count += length * (length + 1) // 2
        
        # Move to the next potential segment
        i = j + 1
        
        # Handle the same for decreasing segments
        if i < n and nums[i - 1] >= nums[i]:
            j = i
            while j < n - 1 and nums[j] > nums[j + 1]:
                j += 1
            
            # Count subarrays in this decreasing segment
            length = j - i + 1
            total_count += length * (length + 1) // 2
            
            i = j + 1
        else:
            i += 1
            
    return total_count

# Example usage
print(count_incremovable_subarrays([1, 2, 3, 4]))  # Output: 10
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