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Longest Alternating Subarray

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Topics:

Arrays

You are given a 0-indexed integer array nums. A subarray s of length m is called alternating if:

m is greater than 1.
s₁ = s₀ + 1.
The 0-indexed subarray s looks like [s₀, s₁, s₀, s₁,...,s_{(m-1) % 2}]. In other words, s₁ - s₀ = 1, s₂ - s₁ = -1, s₃ - s₂ = 1, s₄ - s₃ = -1, and so on up to s[m - 1] - s[m - 2] = (-1)^m.

Return the maximum length of all alternating subarrays present in nums or -1 if no such subarray exists.

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

Example 1:

Input: nums = [2,3,4,3,4]

Output: 4

Explanation:

The alternating subarrays are [2, 3], [3,4], [3,4,3], and [3,4,3,4]. The longest of these is [3,4,3,4], which is of length 4.

Example 2:

Input: nums = [4,5,6]

Output: 2

Explanation:

[4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

Try coding it in LeetCode

Longest Alternating Subarray

Easy

Asked by:

26 views

Topics:

Arrays

You are given a 0-indexed integer array nums. A subarray s of length m is called alternating if:

m is greater than 1.
s₁ = s₀ + 1.
The 0-indexed subarray s looks like [s₀, s₁, s₀, s₁,...,s_{(m-1) % 2}]. In other words, s₁ - s₀ = 1, s₂ - s₁ = -1, s₃ - s₂ = 1, s₄ - s₃ = -1, and so on up to s[m - 1] - s[m - 2] = (-1)^m.

Return the maximum length of all alternating subarrays present in nums or -1 if no such subarray exists.

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

Example 1:

Input: nums = [2,3,4,3,4]

Output: 4

Explanation:

The alternating subarrays are [2, 3], [3,4], [3,4,3], and [3,4,3,4]. The longest of these is [3,4,3,4], which is of length 4.

Example 2:

Input: nums = [4,5,6]

Output: 2

Explanation:

[4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

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 behavior if the input array is null or empty?
Can the input array contain negative numbers, zeros, or floating-point numbers, or is it strictly positive integers?
What defines 'alternating'? Does it mean strictly increasing then decreasing, or decreasing then increasing, or either, and does the difference need to be exactly 1, or can it be any positive number?
If there are multiple subarrays with the same maximum length that satisfy the alternating condition, should I return the first one encountered, or is there a specific rule for choosing among them?
If no alternating subarray exists (e.g., the array is constant or has length 1), what should the function return? Should I return 0, null, or throw an exception?

Brute Force Solution

Approach

The brute force strategy for finding the longest alternating subarray involves checking every possible subarray within the given sequence. It's like examining every conceivable slice of the original data to see if it meets the 'alternating' condition. We will keep track of the longest one that satisfies the rule.

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

Start by looking at the first element alone. Is it an alternating subarray?
Next, consider the first two elements together. Do they alternate?
Then, examine the first three elements, and so on, until you've checked all possible starting points and lengths from the beginning of the sequence.
Now, shift your focus. Begin with the second element alone, then the second and third, then the second, third, and fourth, and so on. Again, check each subarray for the alternating property.
Continue this process, moving your starting point one element at a time, until you've considered every possible subarray within the original sequence.
For each subarray you examine, remember its length if it meets the 'alternating' condition.
Finally, compare the lengths of all the alternating subarrays you've found. The longest one is the answer.

Code Implementation

def longest_alternating_subarray_brute_force(sequence):
    max_length = 0
    sequence_length = len(sequence)

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

            # Subarrays of length 0 or 1 are alternating
            if sub_array_length <= 1:
                if sub_array_length > max_length:
                    max_length = sub_array_length
                continue

            is_alternating = True
            
            # Check if the current sub_array is alternating
            for index in range(sub_array_length - 1):

                # Check adjacent elements to see if they alternate.
                if (sub_array[index] > 0 and sub_array[index + 1] > 0) or \
                   (sub_array[index] < 0 and sub_array[index + 1] < 0):
                    is_alternating = False
                    break

            # Update max_length if necessary
            if is_alternating:
                if sub_array_length > max_length:
                    max_length = sub_array_length

    return max_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. For each starting index i, it considers subarrays of lengths from 1 up to n-i. For each subarray, it needs to check if it's alternating. The outer loop runs n times, and the inner loop, which checks the alternating property and varies based on the subarray length, can also run up to n times in the worst case. This results in approximately n * n/2 operations, which simplifies to O(n²).

Space Complexity

O(1) – The described brute force strategy only requires a few constant space variables to track the start and end indices of the subarray being checked, as well as the length of the longest alternating subarray found so far. No additional data structures that scale with the input size N (the length of the input array) are used. Therefore, the space complexity is constant.

Optimal Solution

Approach

The key is to recognize patterns and reuse information to avoid redundant work. We'll look at each position and extend a potential subarray only as long as it follows the alternating pattern, restarting when the pattern breaks. This way, we only make one 'pass' through the data.

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

Start by looking at the first number.
Check the next number to see if it has the opposite sign (positive/negative) of the current number.
If it does, then extend the current alternating subarray. Keep doing this as long as the signs keep alternating.
If the next number doesn't have the opposite sign, the alternating subarray ends. However, carefully consider if this new number could start a new, longer alternating subarray *right after* the previous one ended.
Keep track of the longest alternating subarray found so far.
Move through all the numbers, extending or restarting alternating subarrays and keeping track of the longest one.
Return the length of the longest alternating subarray that was found.

Code Implementation

def longest_alternating_subarray(numbers):
    max_length = 0
    current_length = 0
    index = 0

    while index < len(numbers):
        current_length = 1
        current_index = index + 1

        # Check for alternating signs
        while current_index < len(numbers) and \
              (numbers[current_index] > 0) != (numbers[current_index - 1] > 0):
            current_length += 1
            current_index += 1

        max_length = max(max_length, current_length)

        # If current_length is 1, move to the next number
        if current_length == 1:
            index += 1
        else:
            # Restart from the second element of current subarray
            index = current_index - current_length + 1

    return max_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once. In each iteration, it checks if the current element's sign alternates with the previous one. If it doesn't, it potentially restarts the subarray from the current element. However, even with restarts, each element is visited at most a constant number of times. Therefore, the time complexity is directly proportional to the number of elements in the input array, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm iterates through the input array, keeping track of the current longest alternating subarray length. It does not create any additional data structures like lists, hash maps, or arrays whose size scales with the input size N. Instead, it uses a few variables to store the current length and the overall maximum length encountered so far, meaning the auxiliary space remains constant irrespective of the input size N.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there's no subarray to evaluate.
Array with a single element	Return 1 as any single element is an alternating subarray of length 1.
Array with only two elements, where they are not alternating.	Return 1, as the longest alternating subarray has length 1 (either the first or second element).
Array with all identical values	Return 1, as no two adjacent elements alternate, so only single-element subarrays are valid.
Array with alternating values from start to end	The solution should correctly identify the entire array as the longest alternating subarray.
Array with alternating values followed by non-alternating values	The algorithm must accurately determine the cutoff point of the alternating sequence.
Integer overflow when calculating differences	Use a data type with a larger range, like `long`, for intermediate calculations or ensure the input array has a limited range to prevent overflow.
Large input array causing time complexity issues	Ensure the algorithm is O(n) or O(n log n) to avoid timeouts with large arrays.

Null or empty input array

How to Handle:

Return 0 immediately as there's no subarray to evaluate.

Array with a single element

How to Handle:

Return 1 as any single element is an alternating subarray of length 1.

Array with only two elements, where they are not alternating.

How to Handle:

Return 1, as the longest alternating subarray has length 1 (either the first or second element).

Array with all identical values

How to Handle:

Return 1, as no two adjacent elements alternate, so only single-element subarrays are valid.

Array with alternating values from start to end

How to Handle:

The solution should correctly identify the entire array as the longest alternating subarray.

Array with alternating values followed by non-alternating values

How to Handle:

The algorithm must accurately determine the cutoff point of the alternating sequence.

Integer overflow when calculating differences

How to Handle:

Use a data type with a larger range, like `long`, for intermediate calculations or ensure the input array has a limited range to prevent overflow.

Large input array causing time complexity issues

How to Handle:

Ensure the algorithm is O(n) or O(n log n) to avoid timeouts with large arrays.

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 behavior if the input array is null or empty?
Can the input array contain negative numbers, zeros, or floating-point numbers, or is it strictly positive integers?
What defines 'alternating'? Does it mean strictly increasing then decreasing, or decreasing then increasing, or either, and does the difference need to be exactly 1, or can it be any positive number?
If there are multiple subarrays with the same maximum length that satisfy the alternating condition, should I return the first one encountered, or is there a specific rule for choosing among them?
If no alternating subarray exists (e.g., the array is constant or has length 1), what should the function return? Should I return 0, null, or throw an exception?

Brute Force Solution

Approach

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

Start by looking at the first element alone. Is it an alternating subarray?
Next, consider the first two elements together. Do they alternate?
Then, examine the first three elements, and so on, until you've checked all possible starting points and lengths from the beginning of the sequence.
Now, shift your focus. Begin with the second element alone, then the second and third, then the second, third, and fourth, and so on. Again, check each subarray for the alternating property.
Continue this process, moving your starting point one element at a time, until you've considered every possible subarray within the original sequence.
For each subarray you examine, remember its length if it meets the 'alternating' condition.
Finally, compare the lengths of all the alternating subarrays you've found. The longest one is the answer.

Code Implementation

def longest_alternating_subarray_brute_force(sequence):
    max_length = 0
    sequence_length = len(sequence)

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

            # Subarrays of length 0 or 1 are alternating
            if sub_array_length <= 1:
                if sub_array_length > max_length:
                    max_length = sub_array_length
                continue

            is_alternating = True
            
            # Check if the current sub_array is alternating
            for index in range(sub_array_length - 1):

                # Check adjacent elements to see if they alternate.
                if (sub_array[index] > 0 and sub_array[index + 1] > 0) or \
                   (sub_array[index] < 0 and sub_array[index + 1] < 0):
                    is_alternating = False
                    break

            # Update max_length if necessary
            if is_alternating:
                if sub_array_length > max_length:
                    max_length = sub_array_length

    return max_length

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by looking at the first number.
Check the next number to see if it has the opposite sign (positive/negative) of the current number.
If it does, then extend the current alternating subarray. Keep doing this as long as the signs keep alternating.
If the next number doesn't have the opposite sign, the alternating subarray ends. However, carefully consider if this new number could start a new, longer alternating subarray *right after* the previous one ended.
Keep track of the longest alternating subarray found so far.
Move through all the numbers, extending or restarting alternating subarrays and keeping track of the longest one.
Return the length of the longest alternating subarray that was found.

Code Implementation

def longest_alternating_subarray(numbers):
    max_length = 0
    current_length = 0
    index = 0

    while index < len(numbers):
        current_length = 1
        current_index = index + 1

        # Check for alternating signs
        while current_index < len(numbers) and \
              (numbers[current_index] > 0) != (numbers[current_index - 1] > 0):
            current_length += 1
            current_index += 1

        max_length = max(max_length, current_length)

        # If current_length is 1, move to the next number
        if current_length == 1:
            index += 1
        else:
            # Restart from the second element of current subarray
            index = current_index - current_length + 1

    return max_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately as there's no subarray to evaluate.
Array with a single element	Return 1 as any single element is an alternating subarray of length 1.
Array with only two elements, where they are not alternating.	Return 1, as the longest alternating subarray has length 1 (either the first or second element).
Array with all identical values	Return 1, as no two adjacent elements alternate, so only single-element subarrays are valid.
Array with alternating values from start to end	The solution should correctly identify the entire array as the longest alternating subarray.
Array with alternating values followed by non-alternating values	The algorithm must accurately determine the cutoff point of the alternating sequence.
Integer overflow when calculating differences	Use a data type with a larger range, like `long`, for intermediate calculations or ensure the input array has a limited range to prevent overflow.
Large input array causing time complexity issues	Ensure the algorithm is O(n) or O(n log n) to avoid timeouts with large arrays.

Null or empty input array

How to Handle:

Return 0 immediately as there's no subarray to evaluate.

Array with a single element

How to Handle:

Return 1 as any single element is an alternating subarray of length 1.

Array with only two elements, where they are not alternating.

How to Handle:

Return 1, as the longest alternating subarray has length 1 (either the first or second element).

Array with all identical values

How to Handle:

Return 1, as no two adjacent elements alternate, so only single-element subarrays are valid.

Array with alternating values from start to end

How to Handle:

The solution should correctly identify the entire array as the longest alternating subarray.

Array with alternating values followed by non-alternating values

How to Handle:

The algorithm must accurately determine the cutoff point of the alternating sequence.

Integer overflow when calculating differences

How to Handle:

Use a data type with a larger range, like `long`, for intermediate calculations or ensure the input array has a limited range to prevent overflow.

Large input array causing time complexity issues

How to Handle:

Ensure the algorithm is O(n) or O(n log n) to avoid timeouts with large arrays.