5 Best Ways to Program to Find Length of Longest Sign Alternating Subsequence from a List of Numbers in Python

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πŸ’‘ Problem Formulation: Finding the length of the longest sign alternating subsequence within a list of numbers entails identifying a sequence where each number alternates in sign (positive to negative or vice versa) compared to the number directly before it. For instance, given the list [1, -2, 3, -4, 5, -6], the longest sign alternating subsequence would be [1, -2, 3, -4, 5, -6] with a length of 6.

Method 1: Dynamic Programming

Dynamic programming is a method wherein we solve complex problems by breaking them down into simpler subproblems. For this problem, dynamic programming can be used to keep track of the longest alternating subsequence ending with a positive and a negative number at each point in the array.

Here’s an example:

def longest_alternating_subsequence(arr):
    n = len(arr)
    las = [[1, 1] for _ in range(n)]
    res = 1
    
    for i in range(1, n):
        for j in range(i):
            if arr[j]  arr[i]:
                las[i][1] = max(las[i][1], las[j][0] + 1)
        res = max(res, max(las[i]))
    
    return res

# Example array
print(longest_alternating_subsequence([1, -2, 3, -4, 5, -6]))

Output: 6

This code snippet defines a function longest_alternating_subsequence() that takes an array of numbers and returns the length of the longest sign alternating subsequence. We use a 2D list las where las[i][0] and las[i][1] respectively store the length of the longest alternating subsequence ending at index i with a positive and negative number. Dynamic programming is applied by iteratively updating the las list at each step using the previous computations.

Method 2: Greedy Approach

A greedy algorithm builds up a solution piece by piece, always choosing the next piece with the most immediate benefit. In this case, the benefit is the alternation in the sign of the subsequences.

Here’s an example:

def longest_alternating_subsequence(arr):
    prev_sign = 0
    len_subseq = 0
    
    for num in arr:
        sign = num > 0
        if sign != prev_sign:
            len_subseq += 1
            prev_sign = sign
    
    return len_subseq

# Example array
print(longest_alternating_subsequence([1, -2, 3, -4, 5, -6]))

Output: 6

The function longest_alternating_subsequence() takes an array and uses a greedy approach to build the longest alternating subsequence. It does this by iterating over the numbers and checking if the current number’s sign is different from the last number’s. If so, it is added to the subsequence and the sign is updated. This method efficiently computes the desired length without needing additional space.

Method 3: Recursive Approach

The recursive approach breaks the problem into smaller instances of the same problem, solving each recursively and utilizing the solutions to solve the larger problem.

Here’s an example:

def util(arr, index, last_sign):
    if index == len(arr):
        return 0
    elif (arr[index]  0 and last_sign == 0):
        return max(util(arr, index + 1, not last_sign) + 1, util(arr, index + 1, last_sign))
    else:
        return util(arr, index + 1, last_sign)

def longest_alternating_subsequence(arr):
    return max(util(arr, 0, 0), util(arr, 0, 1))

# Example array
print(longest_alternating_subsequence([1, -2, 3, -4, 5, -6]))

Output: 6

In this recursive example, we define a utility function util() which is called by longest_alternating_subsequence(). It constructs two recursive branches: one includes the current number if it alternates in sign and one skips it. The base case returns 0 when the end of the array is reached. Although elegantly simple, recursive methods can lead to performance issues due to redundant calculations and risk of stack overflows for large input sizes.

Method 4: Iterative Bottom-Up Approach

The iterative bottom-up approach is similar to dynamic programming but uses iteration rather than recursion to construct solutions to increasingly larger subproblems in a loop.

Here’s an example:

def longest_alternating_subsequence(arr):
    up = down = 1
    
    for i in range(1, len(arr)):
        if arr[i] > arr[i-1]:
            up = down + 1
        elif arr[i] < arr[i-1]:
            down = up + 1
            
    return max(up, down)

# Example array
print(longest_alternating_subsequence([1, -2, 3, -4, 5, -6]))

Output: 6

This code snippet uses an iterative bottom-up approach to solve the problem with two variables up and down, representing the length of the longest alternating subsequence ending with a positive and negative number. Each number in the array is compared with the previous one, and the lengths are updated accordingly. The iterative approach is generally more space-efficient than recursive methods and avoids potential stack overflows.

Bonus One-Liner Method 5: Pythonic Way with itertools

Using Python’s itertools module combined with list comprehensions, we can provide a concise and pythonic one-liner solution for the problem.

Here’s an example:

from itertools import tee

def pairwise(iterable):
    a, b = tee(iterable)
    next(b, None)
    return zip(a, b)

arr = [1, -2, 3, -4, 5, -6]
print(sum(x * y < 0 for x, y in pairwise(arr)) + (len(arr) > 0))

Output: 6

The one-liner utilizes the itertools.tee() function to create two iterators and the pairwise() helper function to iterate through the array in adjacent pairs. The list comprehension checks if consecutive elements have opposite signs, counting how many times this occurs and adds one if the array is not empty. While this method is elegant and pythonic, it may be less clear to those unfamiliar with itertools and Python’s more advanced features.

Summary/Discussion

  • Method 1: Dynamic Programming. Provides an exact and optimized solution. Can be memory-intensive for large inputs.
  • Method 2: Greedy Approach. Offers a simple and efficient solution, but lacks the formal proof of optimality in some cases.
  • Method 3: Recursive Approach. Easy to understand, but not very efficient due to possible overlapping subproblems.
  • Method 4: Iterative Bottom-Up Approach. More space-efficient than recursive solutions. It might be difficult to construct for more complex substructure relationships.
  • Bonus Method 5: Pythonic Way with itertools. Very concise and readable for experienced Python developers, but potentially obscure for beginners.