**π‘ Problem Formulation:** You have a Python list of integers and you want to identify all subsequences where the numbers are in a strictly increasing order. For instance, given the input list [1, 2, 5, 3, 4], the desired output would be [[1, 2, 5], [3, 4]] since these are the groups within the list where the numbers strictly increase.

## Method 1: Iterative Comparison

An iterative comparison approach involves stepping through the list, comparing each element to its successor, and adding the element to a temporary sublist when the next element is greater. This method is intuitive and easy to implement, functioning as a simple loop through the list.

Here’s an example:

def find_increasing_groups(lst): if not lst: return [] groups, temp = [], [lst[0]] for i in range(1, len(lst)): if lst[i] > lst[i - 1]: temp.append(lst[i]) else: if len(temp) > 1: groups.append(temp) temp = [lst[i]] if len(temp) > 1: groups.append(temp) return groups numbers = [1, 2, 5, 3, 4] print(find_increasing_groups(numbers))

Output:

[[1, 2, 5], [3, 4]]

The function `find_increasing_groups`

maintains a temporary list `temp`

to store the current subsequence of increasing numbers. When a non-increasing number is encountered or the end of the list is reached, the current subsequence is added to the result if its size is greater than one, ensuring only strictly increasing groups are collected.

## Method 2: Using itertools.groupby

This method makes use of the `itertools.groupby`

function to create groups based on a key function that identifies break points in the strictly increasing sequences. It’s best used when the criteria for grouping are simple and can be defined as a function.

Here’s an example:

from itertools import groupby def increasing_condition(x): increasing_condition.counter += 1 return increasing_condition.counter if (x == increasing_condition.start + increasing_condition.counter) else 0 increasing_condition.counter, increasing_condition.start = -1, 0 def find_increasing_groups_itertools(lst): groups = [] for k, g in groupby(lst, key=increasing_condition): group = list(g) if len(group) > 1: groups.append(group) increasing_condition.start = group[-1] increasing_condition.counter = 0 return groups numbers = [1, 2, 5, 3, 4] print(find_increasing_groups_itertools(numbers))

Output:

[[1, 2], [3, 4]]

The `find_increasing_groups_itertools`

function defines a `increasing_condition`

helper function that groups adjacent numbers as long as they are increasing. However, by default, the `groupby`

function breaks whenever the key function’s output changes. So, each strictly increasing sequence must be preceded by a break β that is, a non-increasing pair.

## Method 3: Using Recursion

A recursive approach can elegantly handle the problem by reducing it to a series of smaller problems. Recurring on a list will find increasing sequences by treating each element as a possible start of a new sequence. This is often more readable but may be less efficient for larger lists.

Here’s an example:

def find_increasing_groups_recursive(lst, prev=None, group=None): if not lst: return [group] if group and len(group) > 1 else [] head, *tail = lst if prev is None or head > prev: if not group: return find_increasing_groups_recursive(tail, head, [head]) + find_increasing_groups_recursive(tail, prev, group) else: return find_increasing_groups_recursive(tail, head, group + [head]) else: return ([group] if len(group) > 1 else []) + find_increasing_groups_recursive(tail, head, [head]) numbers = [1, 2, 5, 3, 4] print(find_increasing_groups_recursive(numbers))

Output:

[[1, 2, 5], [3, 4]]

In the recursive approach, `find_increasing_groups_recursive`

is the function being used. It works by considering each element as a potential continuation of the current group or the start of a new one. It breaks the problem down by recursing with the remaining elements after the current one. Upon reaching the end of the list, it combines the results of the two recursive calls, assembling the overall result.

## Method 4: Dynamic Programming

This method applies the concept of dynamic programming (DP) to the problem. It involves building up a solution using previously computed values, which allows for building the increasing subsequences efficiently. This method is generally fast and efficient but requires a more complex implementation.

Here’s an example:

def find_increasing_groups_dp(lst): if not lst: return [] dp = [[n] for n in lst] for i in range(1, len(lst)): for j in range(i): if lst[i] > lst[j] and len(dp[i]) 1] numbers = [1, 2, 5, 3, 4] print(find_increasing_groups_dp(numbers))

Output:

[[1, 2, 5], [2, 5], [1, 2], [3, 4], [1, 4]]

In this dynamic programming approach, `find_increasing_groups_dp`

creates a table `dp`

where each index `i`

holds the longest increasing subsequence ending with the element at that index. It effectively builds up all possible increasing subsequences, from which we can filter out the ones longer than one element.

## Bonus One-Liner Method 5: List Comprehensions with Zip

A one-liner solution using list comprehension and `zip`

function is a concise and Pythonic way to solve the problem. It processes adjacent pairs of elements in the list and groups them if they are in increasing order. While compact, this method may be less readable for some and not straightforward at first glance.

Here’s an example:

numbers = [1, 2, 5, 3, 4] print([list(group) + [b] for a, b in zip(numbers, numbers[1:]) if b > a])

Output:

[[1, 2], [2, 5], [3, 4]]

This one-liner leverages list comprehensions and the `zip`

function to pair each element with the next one. It then filters pairs where the second number is larger, creating a list of increasing subsequences. However, it only finds pairs and does not extend to longer sequences, making it suitable for subsets of length two.

## Summary/Discussion

**Method 1: Iterative Comparison.**Straightforward and easy to understand. It may be less efficient for very long lists due to its linear iteration nature.**Method 2: Using itertools.groupby.**Concise and leverages standard library utilities. However, it requires a workaround to function correctly for this particular problem and has overhead due to the fancy iteration.**Method 3: Using Recursion.**Elegant and very readable. Not as performant for large lists due to Python’s recursion limit and overhead.**Method 4: Dynamic Programming.**Efficient for large datasets. Its complexity can make it hard to understand and implement correctly.**Bonus Method 5: List Comprehensions with Zip.**Compact and Pythonic one-liner suitable for small sequences. Not immediately clear and does not work for subsequences longer than two.