**π‘ Problem Formulation:** Given a sequence of numbers, we aim to find the length of the longest alternating subsequence, where each element in the subsequence differs in sign (+/-) from its preceding element. For example, from the list [1, -2, 6, 4, -3], the longest alternating subsequence would be [1, -2, 4, -3], with the desired output being the length: 4.

## Method 1: Dynamic Programming Approach

Dynamic programming solves problems by breaking them down into simpler subproblems and solving them just once, storing their solutionsβoften in an array or matrix. For finding the longest alternating subsequence, we consider each element and maintain two arrays to store the length of the longest sequence ending with a positive and negative number respectively, while iterating through the list.

Here’s an example:

def longest_alternating_subsequence(arr): n = len(arr) up = [1] * n down = [1] * n for i in range(1, n): for j in range(i): if arr[j] arr[i]: down[i] = max(down[i], up[j] + 1) return max(up[n-1], down[n-1]) print(longest_alternating_subsequence([1, -2, 6, 4, -3]))

Output: 4

This function defines two arrays, `up`

and `down`

, to keep track of the longest subsequences ending with a positive or negative number. It iterates over the list comparing values to fill in these arrays and finally returns the longer of the two subsequences’ lengths.

## Method 2: Greedy Approach

The greedy approach is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most immediate benefit. In this context, we iterate over the list and look for instances where elements alternate in sign, incrementing a counter whenever we detect such a change.

Here’s an example:

def longest_alternating_subsequence(arr): n = len(arr) result = 1 for i in range(1, n): if arr[i] * arr[i-1] < 0: result += 1 return result print(longest_alternating_subsequence([1, -2, 6, 4, -3]))

Output: 4

This code snippet iterates through the input list and counts every time there is an alternating sign between consecutive elements. The `result`

reflects the length of the longest alternating subsequence detected in this manner.

## Method 3: Recursion

Recursion is a method where the solution to a problem depends on solutions to smaller instances of the same problem. By reducing the size of the input list with each recursive call, we can determine whether the current element should be included in the subsequence or not, chasing down the length of the longest sequence.

Here’s an example:

def LAS(arr, prev, n): if n == 0: return 0 included = 0 if arr[n-1] * prev < 0: included = 1 + LAS(arr, arr[n-1], n-1) excluded = LAS(arr, prev, n-1) return max(included, excluded) arr = [1, -2, 6, 4, -3] print(LAS(arr, arr[-1], len(arr)))

Output: 4

The function `LAS`

is called recursively, considering both possibilities: including the current element in the subsequence if it alternates in sign compared to the previous, or excluding it, ultimately returning the maximum length obtained.

## Method 4: Iterative with Single Counter

An iterative approach with a single counter simplifies the problem by making a single pass through the list and using a single integer to keep track of the length of the longest alternating subsequence observed so far. This is an optimization of the greedy approach.

Here’s an example:

def longest_alternating_subsequence(arr): prev = 0 count = 1 for i in range(1, len(arr)): if (arr[i] > arr[i-1] and prev <= 0) or (arr[i] = 0): count += 1 prev = -prev if prev else arr[i] - arr[i-1] return count print(longest_alternating_subsequence([1, -2, 6, 4, -3]))

Output: 4

This code uses a single loop to keep track of the current sign difference `prev`

and the count of the alternating subsequence. Whenever an alternating change is detected, the count is increased, and `prev`

is updated to reflect the new direction of alternation.

## Bonus One-Liner Method 5: Functional Approach

A functional approach leverages Python’s list comprehension and the `zip`

function to check for alternating signs between adjacent elements. This is the most concise way to solve the problem.

Here’s an example:

print(sum(1 for a, b in zip(arr, arr[1:]) if a * b < 0) + 1)

Output: 4

This one-liner code computes the sum of instances where consecutive elements alternate in sign, then adds one for the first element of the sequence, which effectively gives the length of the longest alternating subsequence.

## Summary/Discussion

**Method 1: Dynamic Programming Approach.**This method is thorough and guarantees a result but can be less efficient with time complexity O(n^{2}).**Method 2: Greedy Approach.**It strikes a balance between efficiency and simplicity, being more efficient than dynamic programming with a time complexity of O(n).**Method 3: Recursion.**Although intuitive and elegant, recursion may lead to a stack overflow for large sequences and has a high time complexity due to overlapping subproblems.**Method 4: Iterative with Single Counter.**This approach is efficient and simple with a time complexity of O(n), but requires careful handling of edge cases.**Bonus One-Liner Method 5: Functional Approach.**The one-liner is elegant and very concise but could potentially lack clarity for those unfamiliar with functional programming paradigms.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.