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Count Alternating Subarrays

Medium

Count Alternating Subarrays

Medium

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 return value when the input array is null or empty?
Can the input array contain non-integer values?
What is the definition of 'alternating' for this problem? Specifically, does it mean strictly alternating signs (positive, negative, positive...) or just different values (e.g., 1, 2, 1, 2...)?
If multiple alternating subarrays exist, should I return the count of all such subarrays or only the longest one?
What are the constraints on the integer values within the array (e.g., range)? Could there be integer overflow issues I need to consider?

Brute Force Solution

Approach

The basic idea is to check every possible subarray to see if it meets our alternating criteria. We'll consider every possible starting point and then every possible ending point to define our subarray, and then check that subarray.

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

Consider each possible starting position in the list of numbers.
For each starting position, consider every possible ending position that comes after it or is the same as it.
For each of these 'chunks' (subarrays), check if the numbers alternate in sign (positive, negative, positive, etc.).
If the numbers in the chunk alternate in sign, count that chunk as a valid alternating subarray.
After checking every possible chunk, report the total count of valid alternating subarrays we found.

Code Implementation

def count_alternating_subarrays_brute_force(numbers):
    alternating_subarray_count = 0
    list_length = len(numbers)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            # Iterate through all possible subarrays

            is_alternating = True
            if end_index - start_index >= 1:
                for index in range(start_index + 1, end_index + 1):
                    # Check for sign alternation in each subarray

                    if (numbers[index] > 0 and numbers[index - 1] > 0) or \
                            (numbers[index] < 0 and numbers[index - 1] < 0) or \
                            (numbers[index] == 0 and numbers[index - 1] != 0) or \
                            (numbers[index] != 0 and numbers[index - 1] == 0):
                        is_alternating = False
                        break

            # Count the subarray if it meets the criteria
            if is_alternating:
                alternating_subarray_count += 1

    return alternating_subarray_count

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through all possible subarrays. The outer loop considers each of the n elements as a potential starting point. The inner loop then considers each element from the starting point to the end of the array as a potential ending point, resulting in a nested loop structure. For each subarray defined by the starting and ending points, the algorithm checks if the signs of the elements alternate, which requires iterating through the elements of the subarray in the worst case. Therefore, checking the alternating sign property takes O(n) time within the nested loops. This leads to a total time complexity of O(n * n * n) or O(n³).

Space Complexity

O(1) – The provided solution iterates through all possible subarrays using nested loops, but it checks the alternating sign condition directly within the loops. It does not use any auxiliary data structures like temporary lists, hash maps, or recursion. The only memory used is for a few constant-sized variables (loop counters, a counter for valid subarrays) regardless of the input size N (the size of the input list). Therefore, the space complexity is constant.

Optimal Solution

Approach

The best way to count alternating subarrays is to avoid re-checking parts you've already analyzed. We can do this by keeping track of the length of the current alternating sequence and extending it as we go, or restarting if the sequence breaks.

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

Start at the beginning of the sequence.
Check if the first two numbers alternate (are different). If they are, we have an alternating sequence of length 2. If not, this first element cannot be part of any alternating sequence, so we advance one step.
If the first two alternate, look at the next number and see if it continues the alternating pattern.
If it does continue the pattern, increase the length of the current alternating sequence.
If it breaks the pattern, then the previous number is the end of an alternating subarray. Count all the possible alternating subarrays from the first to second element, first to the third element and so on until the last alternating number.
After that, restart the sequence, beginning with the number where the previous sequence stopped.
Continue in this way until you reach the end of the sequence. The total is the total alternating sequences counted along the way.

Code Implementation

def count_alternating_subarrays(sequence):
    array_length = len(sequence)
    total_alternating_subarrays = 0
    current_index = 0

    while current_index < array_length - 1:
        # Check if the current pair alternates
        if sequence[current_index] == sequence[current_index + 1]:
            current_index += 1
            continue

        alternating_sequence_length = 2
        for next_index in range(current_index + 2, array_length):
            # Check if the next element continues the alternating pattern
            if sequence[next_index] == sequence[next_index - 2]:
                break
            alternating_sequence_length += 1

        # Count subarrays if we have an alternating sequence of length 2 or more
        if alternating_sequence_length >= 2:
            total_alternating_subarrays += alternating_sequence_length - 1

        # Advance the current index to the end of the just-counted sequence.
        current_index += alternating_sequence_length - 1

    return total_alternating_subarrays

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input array of size n only once. During this single pass, it extends or restarts the alternating sequence check. The core operation, which is checking if the next element continues the alternating pattern, takes constant time O(1). Therefore, the overall time complexity is driven by the single pass through the array, making it O(n).

Space Complexity

O(1) – The algorithm, as described, doesn't utilize any auxiliary data structures like arrays, hash maps, or recursion. It only needs to maintain a counter for the current length of the alternating sequence. Therefore, the space required remains constant irrespective of the input size N (the length of the sequence), leading to a space complexity of O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since there are no subarrays.
Array with one element	Return 0 since a subarray of at least length 2 is needed.
Array with two identical elements	Return 0 because a subarray of length 2 with identical elements is not alternating.
Array with a long sequence of alternating elements	The algorithm should efficiently count subarrays of different lengths that start at each index of the alternating sequence.
Array with large integer values	Ensure calculations avoid integer overflow by using appropriate data types (e.g., long if necessary).
Array with a very long sequence of same value followed by alternating pattern.	The algorithm should correctly identify the start of the alternating pattern, skipping the same-value sequence.
All elements in the array are distinct.	The algorithm should correctly identify and count all alternating subarrays in this case.
Input array is null	Throw an IllegalArgumentException or return 0 (consistent with empty array), documenting the null-handling behavior.

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 return value when the input array is null or empty?
Can the input array contain non-integer values?
What is the definition of 'alternating' for this problem? Specifically, does it mean strictly alternating signs (positive, negative, positive...) or just different values (e.g., 1, 2, 1, 2...)?
If multiple alternating subarrays exist, should I return the count of all such subarrays or only the longest one?
What are the constraints on the integer values within the array (e.g., range)? Could there be integer overflow issues I need to consider?

Brute Force Solution

Approach

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

Consider each possible starting position in the list of numbers.
For each starting position, consider every possible ending position that comes after it or is the same as it.
For each of these 'chunks' (subarrays), check if the numbers alternate in sign (positive, negative, positive, etc.).
If the numbers in the chunk alternate in sign, count that chunk as a valid alternating subarray.
After checking every possible chunk, report the total count of valid alternating subarrays we found.

Code Implementation

def count_alternating_subarrays_brute_force(numbers):
    alternating_subarray_count = 0
    list_length = len(numbers)

    for start_index in range(list_length):
        for end_index in range(start_index, list_length):
            # Iterate through all possible subarrays

            is_alternating = True
            if end_index - start_index >= 1:
                for index in range(start_index + 1, end_index + 1):
                    # Check for sign alternation in each subarray

                    if (numbers[index] > 0 and numbers[index - 1] > 0) or \
                            (numbers[index] < 0 and numbers[index - 1] < 0) or \
                            (numbers[index] == 0 and numbers[index - 1] != 0) or \
                            (numbers[index] != 0 and numbers[index - 1] == 0):
                        is_alternating = False
                        break

            # Count the subarray if it meets the criteria
            if is_alternating:
                alternating_subarray_count += 1

    return alternating_subarray_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the beginning of the sequence.
Check if the first two numbers alternate (are different). If they are, we have an alternating sequence of length 2. If not, this first element cannot be part of any alternating sequence, so we advance one step.
If the first two alternate, look at the next number and see if it continues the alternating pattern.
If it does continue the pattern, increase the length of the current alternating sequence.
If it breaks the pattern, then the previous number is the end of an alternating subarray. Count all the possible alternating subarrays from the first to second element, first to the third element and so on until the last alternating number.
After that, restart the sequence, beginning with the number where the previous sequence stopped.
Continue in this way until you reach the end of the sequence. The total is the total alternating sequences counted along the way.

Code Implementation

def count_alternating_subarrays(sequence):
    array_length = len(sequence)
    total_alternating_subarrays = 0
    current_index = 0

    while current_index < array_length - 1:
        # Check if the current pair alternates
        if sequence[current_index] == sequence[current_index + 1]:
            current_index += 1
            continue

        alternating_sequence_length = 2
        for next_index in range(current_index + 2, array_length):
            # Check if the next element continues the alternating pattern
            if sequence[next_index] == sequence[next_index - 2]:
                break
            alternating_sequence_length += 1

        # Count subarrays if we have an alternating sequence of length 2 or more
        if alternating_sequence_length >= 2:
            total_alternating_subarrays += alternating_sequence_length - 1

        # Advance the current index to the end of the just-counted sequence.
        current_index += alternating_sequence_length - 1

    return total_alternating_subarrays

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 since there are no subarrays.
Array with one element	Return 0 since a subarray of at least length 2 is needed.
Array with two identical elements	Return 0 because a subarray of length 2 with identical elements is not alternating.
Array with a long sequence of alternating elements	The algorithm should efficiently count subarrays of different lengths that start at each index of the alternating sequence.
Array with large integer values	Ensure calculations avoid integer overflow by using appropriate data types (e.g., long if necessary).
Array with a very long sequence of same value followed by alternating pattern.	The algorithm should correctly identify the start of the alternating pattern, skipping the same-value sequence.
All elements in the array are distinct.	The algorithm should correctly identify and count all alternating subarrays in this case.
Input array is null	Throw an IllegalArgumentException or return 0 (consistent with empty array), documenting the null-handling behavior.