Count the Number of Incremovable Subarrays I

#672 Most AskedEasy

Count the Number of Incremovable Subarrays I #672 Most Asked

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Microsoft

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ArraysTwo Pointers

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the input array `nums`?
Can the elements in `nums` be negative, zero, or non-integer values?
What constitutes an 'incremovable' subarray? Is it sufficient that the *remaining* array be non-decreasing, or are there other conditions?
Are duplicate numbers allowed in the input array, and how do they affect the 'non-decreasing' property of the remaining array after removing a subarray?
If there are multiple incremovable subarrays, do I need to return a specific one, or can I return any valid one?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every possible selection of a subarray to remove. We check if removing each of these subarrays results in the remaining numbers being in non-decreasing order. If so, we count it as a valid removal.

Here's how the algorithm would work step-by-step:

First, consider removing no numbers at all. Check if the original list of numbers is already in non-decreasing order. If it is, count it as one valid way.
Next, consider removing only the first number, then only the second number, and so on, up to removing only the last number. For each of these possibilities, check if the remaining numbers form a non-decreasing sequence. If they do, count them.
Now, consider removing a group of two adjacent numbers, starting from the beginning of the list. Check if the remaining numbers are in non-decreasing order. Count it if it is.
Continue trying every possible group of two adjacent numbers, shifting the group one position to the right each time.
Repeat this process for groups of three adjacent numbers, then four, and so on, until you consider removing almost the entire list (leaving only one number behind).
For each possible removal, always check if the numbers that are left are in non-decreasing order.
Finally, add up the counts of all the valid removals (including removing nothing) to get the total number of incremovable subarrays.

Code Implementation

def count_incremovable_subarrays(numbers):
    list_length = len(numbers)
    count = 0

    # Iterate through all possible subarray lengths
    for subarray_length in range(list_length + 1):
        for start_index in range(list_length - subarray_length + 1):
            end_index = start_index + subarray_length

            # Create a new list with the subarray removed
            modified_numbers = numbers[:start_index] + numbers[end_index:]

            # Check if the modified list is non-decreasing
            is_non_decreasing = True
            for index in range(len(modified_numbers) - 1):
                if modified_numbers[index] > modified_numbers[index + 1]:
                    is_non_decreasing = False
                    break

            # Increment the count if the subarray removal resulted in a non-decreasing list
            if is_non_decreasing:
                count += 1

    return count

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible subarrays to remove. This involves considering subarrays of length 1 up to length n. For each subarray removal, the algorithm checks if the remaining elements are in non-decreasing order. Checking if the remaining elements are non-decreasing takes O(n) time. Since there are O(n^2) possible subarrays to remove, and each removal requires O(n) for checking, the total time complexity is O(n^2 * n) which simplifies to O(n^3).

Space Complexity

O(1) – The provided brute force approach iterates through all possible subarrays and checks if the remaining elements are in non-decreasing order. While the algorithm considers many subarrays, it doesn't explicitly create or store them in auxiliary data structures like lists or hash maps. The checks for non-decreasing order are likely performed in-place or using a few constant-size variables. Therefore, the algorithm's auxiliary space complexity is O(1), indicating constant space usage irrespective of the input size N (the length of the input array).

Optimal Solution

Approach

The goal is to figure out how many sections of a list we can remove so that the remaining list is sorted. The smart way to do this involves checking which parts at the beginning and end of the list can be kept to form a sorted list.

Here's how the algorithm would work step-by-step:

First, check if the entire list is already sorted. If it is, you can remove any section, so the total number of possible sections represents the answer.
If the list isn't already sorted, find the longest sorted part at the beginning of the list.
Also, find the longest sorted part at the end of the list.
Now, try different combinations of keeping the beginning sorted part and the end sorted part, removing everything in between.
For each of these combinations, check if the combined list (beginning + end) is sorted.
If the combined list is sorted, that means we found a section that can be removed. Count it.
Add up all the valid removable sections you found. This total is the answer.

Code Implementation

def count_incremovable_subarrays_i(number_list):
    list_length = len(number_list)
    count = 0

    # Check if the entire list is sorted
    if all(number_list[i] <= number_list[i + 1] for i in range(list_length - 1)):
        return list_length * (list_length + 1) // 2

    for i in range(list_length):
        for j in range(i, list_length):
            # Create a subarray by excluding elements
            new_list = number_list[:i] + number_list[j+1:]

            # Check if the subarray is sorted.
            if len(new_list) <= 1 or all(new_list[k] <= new_list[k+1] for k in range(len(new_list) - 1)):
                count += 1

    return count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm first checks if the entire array of size n is sorted, taking O(n) time. Then, it identifies the longest sorted prefix and suffix, each costing O(n). The core of the algorithm involves iterating through all possible lengths of the prefix (up to n) and for each prefix length, checking all possible lengths of the suffix (up to n). Inside this nested loop, the algorithm checks if the combined prefix and suffix form a sorted array, which takes O(n) time in the worst case. This nested loop structure leads to O(n * n * n) which should simplify to O(n²). The O(n) array copy is performed but happens at most O(n²) times but should not increase beyond the O(n²). Since comparing can be short circuited the worst case is still O(n²).

Space Complexity

O(1) – The described algorithm primarily uses a few integer variables to store indices for the beginning and end sorted portions of the list. No auxiliary data structures like lists, hash maps, or recursion are utilized that scale with the input size, N (the number of elements in the input list). Therefore, the amount of extra space remains constant irrespective of the size of the input list, leading to a space complexity of O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no subarrays to remove.
Input array with a single element	Return 1 since the entire array is incremovable.
Input array with two elements, both equal	Return 3 as removing either element or both creates an incremovable subarray (empty, or single element).
Input array is already strictly increasing	The entire array can be removed, so return n * (n + 1) / 2, where n is the length of the array.
Input array is strictly decreasing	Iterate and check each subarray for 'incremovability' since many subarrays might be invalid.
Input array contains duplicate consecutive elements	Handle duplicates correctly by checking for strictly increasing conditions, allowing equal consecutive elements only if removed.
Large input array (performance consideration)	Ensure the solution has a time complexity of O(n^2) or better to avoid timeouts for large arrays, possibly using dynamic programming.
Input array with all identical elements	All subarrays are incremovable, and the result should be n * (n + 1) / 2, where n is array length.

Empty input array

How to Handle:

Return 0 as there are no subarrays to remove.

Input array with a single element

How to Handle:

Return 1 since the entire array is incremovable.

Input array with two elements, both equal

How to Handle:

Return 3 as removing either element or both creates an incremovable subarray (empty, or single element).

Input array is already strictly increasing

How to Handle:

The entire array can be removed, so return n * (n + 1) / 2, where n is the length of the array.

Input array is strictly decreasing

How to Handle:

Iterate and check each subarray for 'incremovability' since many subarrays might be invalid.

Input array contains duplicate consecutive elements

How to Handle:

Handle duplicates correctly by checking for strictly increasing conditions, allowing equal consecutive elements only if removed.

Large input array (performance consideration)

How to Handle:

Ensure the solution has a time complexity of O(n^2) or better to avoid timeouts for large arrays, possibly using dynamic programming.

Input array with all identical elements

How to Handle:

All subarrays are incremovable, and the result should be n * (n + 1) / 2, where n is array length.