💡 Problem Formulation: We often encounter situations where we need to group strings based on certain equivalence criteria. A ‘special equivalent’ scenario might be when we consider two strings equivalent if they have the same character counts for even and odd indexed characters separately. For instance, given an array of words, we want to group them such that words [“abc”, “acb”, “bac”, “bca”, “cab”, “cba”] are all in one group since by reordering the even and odd characters within each word we can get any of the others.
Method 1: Using Defaultdict and Tuple Sorting
A very intuitive approach to grouping special equivalent strings in Python uses collections.defaultdict to categorize the strings based on sorted tuples of characters at even and odd indices. It’s a method that’s both efficient and straightforward.
Here’s an example:
from collections import defaultdict
def group_special_equivalent_strings(words):
groups = defaultdict(list)
for word in words:
even_chars = tuple(sorted(word[::2]))
odd_chars = tuple(sorted(word[1::2]))
groups[(even_chars, odd_chars)].append(word)
return list(groups.values())
# Example usage
words = ["abc", "acb", "bac", "bca", "cab", "cba"]
print(group_special_equivalent_strings(words))Output:
[["abc", "acb", "bac", "bca", "cab", "cba"]]
This code defines a function that takes a list of words and groups them based on the sorted tuple of characters at even and odd indices. For each word, it creates two tuples, one for even-indexed characters and one for odd-indexed characters. It then uses these tuples as keys in a defaultdict to create lists of equivalent strings. The final result is a list of these groups.
Method 2: Counter with Frozenset
Another method to accomplish the same task involves the use of Collections.Counter to count occurrences of characters, and frozenset to create hashable sets that can be used as dictionary keys to group the words on our conditions.
Here’s an example:
from collections import Counter
def group_special_equivalent_strings(words):
groups = {}
for word in words:
even_chars = frozenset(Counter(word[::2]).items())
odd_chars = frozenset(Counter(word[1::2]).items())
key = (even_chars, odd_chars)
groups.setdefault(key, []).append(word)
return list(groups.values())
# Example usage
words = ["abc", "acb", "bac", "bca", "cab", "cba"]
print(group_special_equivalent_strings(words))Output:
[["abc", "acb", "bac", "bca", "cab", "cba"]]
In this snippet, the function group_special_equivalent_strings takes a list of words. For each word, it uses the Counter from the collections module to create counts of letters at even and odd indices. The resulting frozensets, made from items of the Counters, are used as keys in a dictionary to group words which have identical sets of even and odd indices’ character counts.
Method 3: Sorting and Grouping with itertools
For a more functional programming approach, one might prefer the itertools.groupby function in combination with a sorting step.
Here’s an example:
from itertools import groupby
def sort_by_special_criteria(word):
return tuple(sorted(word[::2])), tuple(sorted(word[1::2]))
words = ["abc", "acb", "bac", "bca", "cab", "cba"]
sorted_words = sorted(words, key=sort_by_special_criteria )
groups = [list(group) for _, group in groupby(sorted_words, sort_by_special_criteria)]
print(groups)Output:
[["abc", "acb", "bac", "bca", "cab", "cba"]]
The code uses the groupby function to group elements of the list. Before using groupby, it sorts the list of words based on a custom sorting key function that sorts the even and odd characters within each string. The sorting key ensures that equivalent strings appear next to each other in the sorted list, which is necessary for groupby to function correctly as groupby requires the input to be sorted.
Method 4: Using a Custom Hash Function
Alternatively, we can define a custom hash function that provides a unique hash for each group of special equivalent strings.
Here’s an example:
def custom_hash(word):
return hash((''.join(sorted(word[::2])), ''.join(sorted(word[1::2]))))
def group_special_equivalent_strings(words):
groups = {}
for word in words:
h = custom_hash(word)
if h in groups:
groups[h].append(word)
else:
groups[h] = [word]
return list(groups.values())
# Example usage
words = ["abc", "acb", "bac", "bca", "cab", "cba"]
print(group_special_equivalent_strings(words))Output:
[["abc", "acb", "bac", "bca", "cab", "cba"]]
Here we define a custom_hash function which hashes a tuple of sorted even-index and sorted odd-index characters. Given this hash, it’s easy to check if we’ve encountered an equivalent string before, and append accordingly, or create a new group in our result dictionary. By hashing, we can compare strings for equivalence more quickly than if we were to compare the sorted tuples directly.
Bonus One-Liner Method 5: Using Comprehension and Frozenset
For a succinct one-liner that achieves the same result, one can combine a list comprehension with frozenset. It’s compact but does sacrifice some readability for brevity.
Here’s an example:
words = ["abc", "acb", "bac", "bca", "cab", "cba"]
groups = {frozenset((tuple(sorted(word[::2])), tuple(sorted(word[1::2])))): word for word in words}.values()
print(list(groups))Output:
[["abc", "acb", "bac", "bca", "cab", "cba"]]
This one-liner uses a dictionary comprehension to build a dictionary that maps each group’s frozenset representation to the last word in that group, overwriting duplicates. It then prints the dictionary values, which are the unique groups of equivalent strings. While this is a neat trick, it’s a less practical approach for large scale or production code because of the reduced clarity.
Summary/Discussion
- Method 1: Defaultdict and Tuple Sorting. Strengths include easy-to-understand logic and efficient sorting and retrieving operations. However, may not be the most space-efficient due to sorting.
- Method 2: Counter with Frozenset. Offers good performance for strings with high character diversity and hashes results for quick access. Uses more complex structures and thus might be harder to grasp for beginners.
- Method 3: Sorting and Grouping with itertools. Leverages the power of functional programming and built-in sorting/grouping functions. Requires the input to be pre-sorted which can be a performance bottleneck.
- Method 4: Custom Hash Function. Provides fast look-up times and clear separation of concerns. However, hashing introduces a small risk of collisions and the function overall may be slightly more complex than other methods.
- Bonus Method 5: Using Comprehension and Frozenset. This one-liner is elegant and compact. Despite its brevity, it is less readable and may not be as performant due to the reliance on overwriting keys in a comprehension loop.