5 Ways to Check Whether a List Can Be Split into Consecutive Increasing Sublists in Python

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πŸ’‘ Problem Formulation: The task is to determine if a given list of integers can be partitioned into disjoint sublists where each sublist is in strictly increasing consecutive order. For instance, given the list [1, 2, 3, 6, 7, 8], it can be split into sublists [1, 2, 3] and [6, 7, 8], which adhere to the required conditions, hence the result would be True.

Method 1: Iterative Approach

This method involves iterating through the list and checking if the difference between consecutive elements is more than 1. If such a difference is found, that position can potentially be the splitting point between sublists. The function should verify if each segment forms a consecutive sequence.

Here’s an example:

def can_split_into_consecutive_sublists(lst):
    start = 0
    for i in range(1, len(lst)):
        if lst[i] - lst[i-1] != 1:
            if i - start <= 1:
                return False
            start = i
    return True
# Input list
input_list = [1, 2, 3, 5, 6, 7]

Output: False

The function can_split_into_consecutive_sublists takes a list as an argument and iterates through it to find discontinuities. In this case, it identifies that the list cannot be split into consecutive increasing sublists since there’s a jump from 3 to 5, without 4, and the sublist [5, 6, 7] doesn’t start from the split point.

Method 2: Using GroupBy from itertools

The groupby function from Python’s itertools module can group consecutive elements. This method uses groupby to cluster the increasing sequences and then checks if the sublists formed by these groups are consecutive and increasing.

Here’s an example:

from itertools import groupby
from operator import itemgetter

def can_be_split(lst):
    for k, g in groupby(enumerate(lst), lambda x: x[1] - x[0]):
        group = list(map(itemgetter(1), g))
        if len(group) < 2 or group != list(range(group[0], group[-1]+1)):
            return False
    return True
input_list = [1, 2, 3, 6, 7, 8]

Output: True

In this snippet, groupby clusters items by the difference between the item’s value and its index, which helps in finding consecutive elements. The result is checked to ensure each group forms a consecutive increasing sublist. The sample list can indeed be split as required, hence the output is True.

Method 3: Using a Custom Generator Function

A custom generator function can be created to yield consecutive increasing sublists. This approach checks if the entire list has been accounted for in these sublists successfully.

Here’s an example:

def consecutive_sublists(lst):
    if not lst: 
    sublist = [lst[0]]
    for item in lst[1:]:
        if item == sublist[-1] + 1:
            yield sublist
            sublist = [item]
    yield sublist

def can_be_split(lst):
    return all(len(sublist) > 1 for sublist in consecutive_sublists(lst))

input_list = [1, 2, 3, 4, 6, 7, 8]

Output: True

The function consecutive_sublists yields increasing consecutive sublists. The main function, can_be_split, then checks if all sublists contain more than one element, ensuring that each sublist is a valid consecutive sequence. The example list shows a valid case, hence resulting in True.

Method 4: Using Recursion

Recursion can be applied to this problem by continually splitting the list into head and tail, checking for the consecutive property in head before proceeding to the tail. It’s an elegant mathematical approach but can be less efficient for long lists due to stack overflow risks.

Here’s an example:

def is_consecutive_head(lst):
    return all(lst[i] + 1 == lst[i + 1] for i in range(len(lst) - 1))

def can_be_split_recursively(lst):
    if len(lst) < 2:
        return False
    for i in range(1, len(lst)):
        if is_consecutive_head(lst[:i]) and can_be_split_recursively(lst[i:]):
            return True
    return False

input_list = [1, 2, 4, 5, 7, 8]

Output: False

The recursion-based function attempts to split the list at every index and checks if the head is a consecutive sequence before recursively checking the tail. It stops when a valid split is found or if it reaches the end of the list. In the given example, the list does not satisfy the condition to split into increasing sublists since it lacks a sequence starting immediately after 2.

Bonus One-Liner Method 5: Using List Comprehension and zip

List comprehension combined with the zip function provides a concise way to check for consecutive elements. This one-liner is pythonic and efficient for short lists but might lack readability for beginners.

Here’s an example:

def can_be_split_oneliner(lst):
    return all(x + 1 == y for x, y in zip(lst, lst[1:]))

input_list = [1, 2, 3, 5, 6, 7, 8]

Output: False

The one-liner uses the zip function to create pairs of consecutive elements from the list and checks if each element is one less than the next. The example list cannot be split into consecutive increasing sublists, as there is no 4 to connect the first and second parts of the list.


  • Method 1: Iterative Approach. Straightforward and easy to understand. However, it might not be the most elegant or the most efficient for very long lists.
  • Method 2: Using GroupBy from itertools. Pythonic and utilizes built-in functions efficiently. Complexity may be difficult for beginners to grasp.
  • Method 3: Using a Custom Generator Function. Modular and reusable. It incurs the overhead of multiple function calls, which may be less efficient than a single-pass approach.
  • Method 4: Using Recursion. Mathematically elegant and clean. Risk of stack overflow with large lists and potentially slow for large inputs due to redundant calculations.
  • Bonus Method 5: One-Liner with List Comprehension and zip. Compact and efficient for small to medium-sized lists. May sacrifice some readability for the sake of brevity.