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Adjacent Increasing Subarrays Detection I

Easy

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Topics:

Arrays

Given an array nums of n integers and an integer k, determine whether there exist two adjacent subarrays of length k such that both subarrays are strictly increasing. Specifically, check if there are two subarrays 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 true if it is possible to find two such subarrays, and false otherwise.

Example 1:

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

Output: true

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, so the result is true.

Example 2:

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

Output: false

Constraints:

2 <= nums.length <= 100
1 < 2 * k <= nums.length
-1000 <= nums[i] <= 1000

Try coding it in LeetCode

Adjacent Increasing Subarrays Detection I

Easy

Asked by:

29 views

Topics:

Arrays

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 true if it is possible to find two such subarrays, and false otherwise.

Example 1:

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

Output: true

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, so the result is true.

Example 2:

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

Output: false

Constraints:

2 <= nums.length <= 100
1 < 2 * k <= nums.length
-1000 <= nums[i] <= 1000

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:

Can the input array contain negative numbers, zeros, or floating-point values?
What should I return if there are no adjacent increasing subarrays in the input?
By 'adjacent,' do you mean consecutive elements in the array, or elements that are next to each other after a potential sorting?
What is the expected output if there are overlapping adjacent increasing subarrays? For example, in `[1, 2, 3, 4]`, should `[1, 2]` and `[2, 3]` both be counted?
Is there a minimum length requirement for the increasing subarrays? For instance, must they contain at least two elements?

Brute Force Solution

Approach

The simplest way to find increasing groups next to each other is to just look at all the groups. We check every possible starting point and every possible size for a group, and then see if it fits our rule about increasing numbers.

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

Start at the very beginning of the list of numbers.
Imagine a group that starts there and only includes the first number. Check if it is an increasing group.
Now imagine a group that starts there and includes the first two numbers. Check if it is an increasing group.
Keep doing this, making the group bigger and bigger, until it reaches the end of the list or is no longer an increasing group.
Once you've checked all the groups that start at the beginning, move to the second number in the list.
Repeat the same process: imagine a group starting there with just one number, then two numbers, and so on, each time checking if it is an increasing group.
Keep going through the list, checking all possible groups starting at each number.
Whenever you find two increasing groups next to each other, make a note of it.
In the end, count how many times you found two increasing groups right next to each other.

Code Implementation

def adjacent_increasing_subarrays_detection_i(numbers):
    number_of_adjacent_increasing_subarrays = 0
    list_length = len(numbers)

    for first_subarray_start in range(list_length):
        for first_subarray_length in range(1, list_length - first_subarray_start + 1):
            first_subarray_end = first_subarray_start + first_subarray_length
            first_subarray = numbers[first_subarray_start:first_subarray_end]

            is_first_subarray_increasing = True
            for index in range(len(first_subarray) - 1):
                if first_subarray[index] >= first_subarray[index + 1]:
                    is_first_subarray_increasing = False
                    break

            if is_first_subarray_increasing:
                # Check for a second increasing subarray immediately after the first

                for second_subarray_length in range(1, list_length - first_subarray_end + 1):
                    second_subarray_start = first_subarray_end
                    second_subarray_end = second_subarray_start + second_subarray_length
                    second_subarray = numbers[second_subarray_start:second_subarray_end]

                    is_second_subarray_increasing = True
                    for index in range(len(second_subarray) - 1):
                        if second_subarray[index] >= second_subarray[index + 1]:
                            is_second_subarray_increasing = False
                            break

                    if is_second_subarray_increasing:
                        # Count the adjacent increasing subarrays
                        number_of_adjacent_increasing_subarrays += 1

                        # The second subarray has been found, no need to continue this loop
                        break

    return number_of_adjacent_increasing_subarrays

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through all possible subarrays within the input array of size n. For each starting index i, it considers all possible lengths of subarrays starting from i. Checking if a subarray is increasing requires iterating through the elements of that subarray. Thus, there are three nested loops: one for the starting index, one for the length of the subarray, and one to check for increasing order inside the identified subarray. The total number of operations approximates n * n * n, which simplifies to O(n³).

Space Complexity

O(1) – The algorithm iterates through the input list, checking subarrays. It doesn't use any auxiliary data structures like lists, hash maps, or stacks to store intermediate results. It only needs a few constant extra variables to track the start and end indices of the current subarrays and a count variable for the adjacent increasing subarrays. Thus, the space used is constant regardless of the input size N.

Optimal Solution

Approach

The core idea is to walk through the sequence once, keeping track of the current increasing streak. We reset the streak counter whenever the sequence dips, looking for contiguous increasing segments.

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

Start at the beginning of the sequence.
Check if the current number is bigger than the one before it. If it is, increase the length of our current increasing segment.
If the current number is not bigger than the previous one, then we've reached the end of an increasing segment. Start counting a new one from this number.
Every time you complete an increasing segment, record its length.
Repeat this process until you reach the end of the sequence.
Return the count of the lengths of the increasing segments that were found.

Code Implementation

def find_adjacent_increasing_subarrays(sequence):
    increasing_segment_lengths = []
    current_segment_length = 0

    # Iterate through the sequence to identify increasing segments
    for i in range(len(sequence)):
        if i == 0:
            current_segment_length = 1
        else:
            # Check if the current element continues the increasing segment
            if sequence[i] > sequence[i - 1]:
                current_segment_length += 1
            else:
                # An increasing segment has ended, so save its length
                increasing_segment_lengths.append(current_segment_length)
                current_segment_length = 1

    # Append the length of the last increasing segment
    increasing_segment_lengths.append(current_segment_length)

    return len(increasing_segment_lengths)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input sequence of size n exactly once. In each iteration, it performs a constant amount of work: comparing the current element to the previous one, updating the streak length, or resetting the streak. Therefore, the time complexity is directly proportional to the size of the input, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the current increasing segment length and a counter for the number of increasing segments. No auxiliary data structures like lists or hashmaps are used to store intermediate results or track visited locations. The space required does not depend on the size of the input sequence, denoted as N. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list immediately, as there are no subarrays to evaluate.
Input array with only one element	Return an empty list, as an increasing subarray requires at least two elements.
Input array with all identical elements	Return an empty list, as no increasing subarray can be formed.
Input array with elements in strictly decreasing order	Return an empty list, as no increasing subarray exists.
Large input array to test for time complexity	Ensure the algorithm's time complexity is linear O(n) to handle large inputs efficiently without timeout.
Input array with negative numbers and/or zeros	The comparison logic should correctly handle negative numbers and zeros as valid increasing elements.
Input array with extreme integer values (Integer.MAX_VALUE, Integer.MIN_VALUE)	Ensure the comparison of adjacent elements does not result in integer overflow issues during the increase check.
Input array with consecutive identical elements followed by an increasing element	The start index should correctly advance past the identical elements before considering the increasing element.

Null or empty input array

How to Handle:

Return an empty list immediately, as there are no subarrays to evaluate.

Input array with only one element

How to Handle:

Return an empty list, as an increasing subarray requires at least two elements.

Input array with all identical elements

How to Handle:

Return an empty list, as no increasing subarray can be formed.

Input array with elements in strictly decreasing order

How to Handle:

Return an empty list, as no increasing subarray exists.

Large input array to test for time complexity

How to Handle:

Ensure the algorithm's time complexity is linear O(n) to handle large inputs efficiently without timeout.

Input array with negative numbers and/or zeros

How to Handle:

The comparison logic should correctly handle negative numbers and zeros as valid increasing elements.

Input array with extreme integer values (Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

Ensure the comparison of adjacent elements does not result in integer overflow issues during the increase check.

Input array with consecutive identical elements followed by an increasing element

How to Handle:

The start index should correctly advance past the identical elements before considering the increasing element.

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:

Can the input array contain negative numbers, zeros, or floating-point values?
What should I return if there are no adjacent increasing subarrays in the input?
By 'adjacent,' do you mean consecutive elements in the array, or elements that are next to each other after a potential sorting?
What is the expected output if there are overlapping adjacent increasing subarrays? For example, in `[1, 2, 3, 4]`, should `[1, 2]` and `[2, 3]` both be counted?
Is there a minimum length requirement for the increasing subarrays? For instance, must they contain at least two elements?

Brute Force Solution

Approach

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

Start at the very beginning of the list of numbers.
Imagine a group that starts there and only includes the first number. Check if it is an increasing group.
Now imagine a group that starts there and includes the first two numbers. Check if it is an increasing group.
Keep doing this, making the group bigger and bigger, until it reaches the end of the list or is no longer an increasing group.
Once you've checked all the groups that start at the beginning, move to the second number in the list.
Repeat the same process: imagine a group starting there with just one number, then two numbers, and so on, each time checking if it is an increasing group.
Keep going through the list, checking all possible groups starting at each number.
Whenever you find two increasing groups next to each other, make a note of it.
In the end, count how many times you found two increasing groups right next to each other.

Code Implementation

def adjacent_increasing_subarrays_detection_i(numbers):
    number_of_adjacent_increasing_subarrays = 0
    list_length = len(numbers)

    for first_subarray_start in range(list_length):
        for first_subarray_length in range(1, list_length - first_subarray_start + 1):
            first_subarray_end = first_subarray_start + first_subarray_length
            first_subarray = numbers[first_subarray_start:first_subarray_end]

            is_first_subarray_increasing = True
            for index in range(len(first_subarray) - 1):
                if first_subarray[index] >= first_subarray[index + 1]:
                    is_first_subarray_increasing = False
                    break

            if is_first_subarray_increasing:
                # Check for a second increasing subarray immediately after the first

                for second_subarray_length in range(1, list_length - first_subarray_end + 1):
                    second_subarray_start = first_subarray_end
                    second_subarray_end = second_subarray_start + second_subarray_length
                    second_subarray = numbers[second_subarray_start:second_subarray_end]

                    is_second_subarray_increasing = True
                    for index in range(len(second_subarray) - 1):
                        if second_subarray[index] >= second_subarray[index + 1]:
                            is_second_subarray_increasing = False
                            break

                    if is_second_subarray_increasing:
                        # Count the adjacent increasing subarrays
                        number_of_adjacent_increasing_subarrays += 1

                        # The second subarray has been found, no need to continue this loop
                        break

    return number_of_adjacent_increasing_subarrays

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The core idea is to walk through the sequence once, keeping track of the current increasing streak. We reset the streak counter whenever the sequence dips, looking for contiguous increasing segments.

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

Start at the beginning of the sequence.
Check if the current number is bigger than the one before it. If it is, increase the length of our current increasing segment.
If the current number is not bigger than the previous one, then we've reached the end of an increasing segment. Start counting a new one from this number.
Every time you complete an increasing segment, record its length.
Repeat this process until you reach the end of the sequence.
Return the count of the lengths of the increasing segments that were found.

Code Implementation

def find_adjacent_increasing_subarrays(sequence):
    increasing_segment_lengths = []
    current_segment_length = 0

    # Iterate through the sequence to identify increasing segments
    for i in range(len(sequence)):
        if i == 0:
            current_segment_length = 1
        else:
            # Check if the current element continues the increasing segment
            if sequence[i] > sequence[i - 1]:
                current_segment_length += 1
            else:
                # An increasing segment has ended, so save its length
                increasing_segment_lengths.append(current_segment_length)
                current_segment_length = 1

    # Append the length of the last increasing segment
    increasing_segment_lengths.append(current_segment_length)

    return len(increasing_segment_lengths)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list immediately, as there are no subarrays to evaluate.
Input array with only one element	Return an empty list, as an increasing subarray requires at least two elements.
Input array with all identical elements	Return an empty list, as no increasing subarray can be formed.
Input array with elements in strictly decreasing order	Return an empty list, as no increasing subarray exists.
Large input array to test for time complexity	Ensure the algorithm's time complexity is linear O(n) to handle large inputs efficiently without timeout.
Input array with negative numbers and/or zeros	The comparison logic should correctly handle negative numbers and zeros as valid increasing elements.
Input array with extreme integer values (Integer.MAX_VALUE, Integer.MIN_VALUE)	Ensure the comparison of adjacent elements does not result in integer overflow issues during the increase check.
Input array with consecutive identical elements followed by an increasing element	The start index should correctly advance past the identical elements before considering the increasing element.

Null or empty input array

How to Handle:

Return an empty list immediately, as there are no subarrays to evaluate.

Input array with only one element

How to Handle:

Return an empty list, as an increasing subarray requires at least two elements.

Input array with all identical elements

How to Handle:

Return an empty list, as no increasing subarray can be formed.

Input array with elements in strictly decreasing order

How to Handle:

Return an empty list, as no increasing subarray exists.

Large input array to test for time complexity

How to Handle:

Ensure the algorithm's time complexity is linear O(n) to handle large inputs efficiently without timeout.

Input array with negative numbers and/or zeros

How to Handle:

The comparison logic should correctly handle negative numbers and zeros as valid increasing elements.

Input array with extreme integer values (Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

Ensure the comparison of adjacent elements does not result in integer overflow issues during the increase check.

Input array with consecutive identical elements followed by an increasing element

How to Handle:

The start index should correctly advance past the identical elements before considering the increasing element.