Adjacent Increasing Subarrays Detection II

Medium

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Arrays

Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where:

Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.
The subarrays must be adjacent, meaning b = a + k.

Return the maximum possible value of k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,5,7,8,9,2,3,4,3,1]

Output: 3

Explanation:

The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Example 2:

Input: nums = [1,2,3,4,4,4,4,5,6,7]

Output: 2

Explanation:

The subarray starting at index 0 is [1, 2], which is strictly increasing.
The subarray starting at index 2 is [3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Constraints:

2 <= nums.length <= 2 * 10⁵
-10⁹ <= nums[i] <= 10⁹

Try coding it in LeetCode

Adjacent Increasing Subarrays Detection II

Medium

Asked by:

7 views

Topics:

Arrays

Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where:

Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.
The subarrays must be adjacent, meaning b = a + k.

Return the maximum possible value of k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,5,7,8,9,2,3,4,3,1]

Output: 3

Explanation:

The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Example 2:

Input: nums = [1,2,3,4,4,4,4,5,6,7]

Output: 2

Explanation:

The subarray starting at index 0 is [1, 2], which is strictly increasing.
The subarray starting at index 2 is [3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Constraints:

2 <= nums.length <= 2 * 10⁵
-10⁹ <= nums[i] <= 10⁹

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 is the expected data type of the array elements (e.g., integers, floats)? Can the array contain negative numbers, zeros, or null values?
What should be returned if no adjacent increasing subarrays are found in the given array? Should I return null, an empty array, or a specific error code?
By 'adjacent increasing subarrays', do you mean that the two subarrays must be immediately next to each other in the input array without any intervening elements?
If there are multiple pairs of adjacent increasing subarrays, is there a specific criteria for which pair to return (e.g., the pair with the longest subarrays, the first pair encountered)?
Can you provide an example of what an adjacent increasing subarray would look like given an example input?

Brute Force Solution

Approach

The brute force method for this problem involves checking every possible consecutive group of numbers in a list to see if they are increasing. We essentially look at every possible chunk of the list and test it. If we find one that fits the rules, we mark it.

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

Start by looking at the first two numbers in the list. Are they in increasing order? If so, remember that.
Then, look at the first three numbers. Are they in increasing order? If so, remember that as well.
Keep doing this, looking at the first four numbers, then five, and so on, each time checking if the numbers are increasing and remembering it if they are.
Once you've checked all the groups of numbers starting from the beginning, move on to the second number in the list.
Repeat the process: look at the second and third numbers, then the second, third, and fourth, and so on, each time checking for increasing order.
Keep moving down the list, starting at each number in turn, and checking all possible increasing groups of numbers that begin there.
When you've gone through the whole list in this way, you'll have checked every possible group of consecutive numbers for increasing order. Count how many of these groups were increasing, and that's your answer.

Code Implementation

def count_adjacent_increasing_subarrays_brute_force(number_list):
    subarray_count = 0

    for start_index in range(len(number_list)): 
        for end_index in range(start_index + 1, len(number_list)): 
            
            sub_array = number_list[start_index:end_index+1]

            # Need at least two elements to form a increasing subarray
            if len(sub_array) < 2:
                continue

            is_increasing = True

            # Check if the subarray is strictly increasing
            for index in range(len(sub_array) - 1):
                if sub_array[index] >= sub_array[index + 1]:
                    is_increasing = False
                    break

            # Increment count if the subarray is increasing
            if is_increasing:
                subarray_count += 1

    return subarray_count

Big(O) Analysis

Time Complexity

O(n²) – The outer loop iterates through each of the n elements in the input array. The inner loop, for each element, checks subarrays of increasing lengths starting from that element. In the worst case, for an element at index i, it checks (n-i) elements. Thus, the total number of checks is approximately n + (n-1) + (n-2) + ... + 1, which is the sum of an arithmetic series. This sum is approximately n * (n+1) / 2, simplifying to O(n²).

Space Complexity

O(1) – The provided algorithm does not explicitly use any auxiliary data structures like lists, hash maps, or sets. It iterates through subarrays, comparing adjacent elements directly within the input list. The only extra memory used would be for a few counter variables to track the start and end of the subarrays and the count of increasing subarrays, which takes up a constant amount of space irrespective of the input size N, where N is the length of the input list. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The best way to solve this is to keep track of how long the increasing sequences are as you move through the numbers. Whenever you find a sequence of the minimum length, you count it and then advance your position past that sequence.

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

Start looking at the numbers one by one from the beginning.
As long as you see each number being bigger than the one before it, keep counting how long the increasing sequence is.
If you encounter a number that isn't bigger than the one before it, check if your sequence is at least as long as the minimum length requirement.
If the increasing sequence is long enough, increase the count of valid sequences by one.
Advance past the just-counted increasing sequence so that you're looking at the numbers after it, regardless of whether the current number continues the sequence or not.
If the increasing sequence is too short, move one number forward to start looking for a new potential sequence.
Repeat these steps until you have examined all the numbers.

Code Implementation

def find_number_of_adjacent_increasing_subarrays(
    numbers, minimum_length
):
    array_length = len(numbers)
    number_of_subarrays = 0
    index = 0

    while index < array_length:
        increasing_sequence_length = 1

        # Count length of current increasing sequence
        while (
            index + increasing_sequence_length < array_length
            and numbers[index + increasing_sequence_length] > numbers[index + increasing_sequence_length - 1]
        ):
            increasing_sequence_length += 1

        # Check if sequence meets length criteria
        if increasing_sequence_length >= minimum_length:
            number_of_subarrays += 1

            # Advance index past counted sequence
            index += increasing_sequence_length

        else:
            # Move one step forward if sequence too short
            index += 1

    return number_of_subarrays

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n. While an increasing sequence is found, the algorithm advances past the entire sequence *at once* when the sequence meets the length requirement. Even if an increasing sequence is not long enough, the algorithm only advances by one position. Since each element is visited and considered for inclusion in a valid sequence at most once, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm iterates through the input array, keeping track of the length of increasing sequences and the count of valid sequences. It uses a few constant-size variables such as the current sequence length and the total valid sequence count. No auxiliary data structures that scale with the input size N (where N is the length of the input array) are used. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array	Return an empty list since no adjacent increasing subarrays can exist.
Array with only one element	Return an empty list since an adjacent increasing subarray requires at least two elements.
Array with two elements, both equal	Return an empty list as the elements are not strictly increasing.
Array with two elements, strictly increasing	Return a list containing the subarray indices [0, 1].
Array with all identical values	Return an empty list as no adjacent elements are strictly increasing.
Array with a long sequence of decreasing values	The algorithm should iterate through the array efficiently, not getting stuck in a complex recursion, correctly skipping over the decreasing sequence and continuing evaluation where appropriate.
Array with very large integers (potential overflow)	Ensure that the comparison of adjacent elements doesn't lead to integer overflow; consider using larger data types if necessary, or alternative comparison logic if overflow is possible given language limitations.
Very large array size exceeding memory limits	Consider processing the array in chunks or using an iterative approach with constant space to avoid memory overflow.