💡 Problem Formulation: Prefix compression involves reducing two input strings to their common beginning, followed by the two unique remainders. Given strings ‘string1’ and ‘string2’, the goal is to identify the common prefix and output a tuple: (length_of_common_prefix, unique_part_of_string1, unique_part_of_string2). For example, for ‘interspecies’ and ‘interstellar’, the desired output would be (5, ‘species’, ‘stellar’).
Method 1: Iterative Comparison
This method involves iteratively comparing characters from the beginning of both strings, character by character, to find the common prefix. It’s a straightforward approach that doesn’t require any external libraries. The function will return a tuple with the length of the common prefix and the unique suffixes of both strings.
Here’s an example:
def prefix_compression(s1, s2):
min_length = min(len(s1), len(s2))
for i in range(min_length):
if s1[i] != s2[i]:
return (i, s1[i:], s2[i:])
return (min_length, s1[min_length:], s2[min_length:])
# Test the function
result = prefix_compression('interspecies', 'interstellar')
print(result)Output:
(5, 'species', 'stellar')
This code snippet defines the function prefix_compression() that iterates through each pair of characters in the input strings and stops when it finds the first non-matching pair. It returns the common prefix length and the unique parts of both strings as a tuple.
Method 2: Using the zip() Function
Python’s built-in zip() function can be used for pairing up elements of two iterables and iterating over these pairs simultaneously. This is an elegant and Pythonic way to compare two strings for prefix compression.
Here’s an example:
def prefix_compression_zip(s1, s2):
i = 0
for a, b in zip(s1, s2):
if a != b:
break
i += 1
return (i, s1[i:], s2[i:])
# Test the function
result = prefix_compression_zip('interspecies', 'interstellar')
print(result)Output:
(5, 'species', 'stellar')
In this code, the prefix_compression_zip() function uses zip() to traverse the characters of both strings in parallel and breaks when the characters do not match. It then returns the prefix length and the remaining parts of the strings.
Method 3: Using Regular Expressions
We can employ regular expressions to quickly find the longest common prefix. Although this method is powerful, it comes with the overhead of processing a regular expression, which can be slower for short strings or unnecessary if the strings are known to be well behaved.
Here’s an example:
import re
def prefix_compression_regex(s1, s2):
match = re.match(r"(.*?)(?=([^a-zA-Z]|$))", s1 + ' ' + s2)
prefix = match.group(0) if match else ''
return (len(prefix), s1[len(prefix):], s2[len(prefix):])
# Test the function
result = prefix_compression_regex('interspecies', 'interstellar')
print(result)Output:
(5, 'species', 'stellar')
The prefix_compression_regex() function finds the common prefix using a regular expression and returns parts of the strings after the common prefix. Note that this function assumes the strings only contain alphanumeric characters.
Method 4: Using os.path.commonprefix
The os.path.commonprefix() function is a convenient function designed originally for finding common path prefixes. It works with any sequence of strings and can be repurposed for our task too.
Here’s an example:
import os
def prefix_compression_os(s1, s2):
prefix = os.path.commonprefix([s1, s2])
return (len(prefix), s1[len(prefix):], s2[len(prefix):])
# Test the function
result = prefix_compression_os('interspecies', 'interstellar')
print(result)Output:
(5, 'species', 'stellar')
The prefix_compression_os() function utilizes os.path.commonprefix() to compute the common prefix length and then slices the input strings accordingly to return the differentiating suffixes alongside the common prefix length.
Bonus One-Liner Method 5: List Comprehension With Enumerate
For a concise one-liner approach, we can combine list comprehension with enumerate() to find the common prefix in a single statement.
Here’s an example:
prefix_compression_one_liner = lambda s1, s2: next(((i, s1[i:], s2[i:]) for i, (a, b) in enumerate(zip(s1, s2)) if a != b), (min(len(s1), len(s2)), s1[len(s2):], s2[len(s1):]))
# Test the function
result = prefix_compression_one_liner('interspecies', 'interstellar')
print(result)Output:
(5, 'species', 'stellar')
This one-liner uses list comprehension and enumerate() alongside a lambda function to yield the first pair of non-matching characters or the complete length of the shorter string, followed by the respective unique suffixes.
Summary/Discussion
- Method 1: Iterative Comparison. Appropriate for beginners and clear in operation. It can be a bit slower if strings are very long due to explicit looping.
- Method 2: Using the zip() Function. Pythonic with clean syntax. However, similar to the iterative approach, its efficiency decreases with longer strings.
- Method 3: Using Regular Expressions. Offers pattern-matching capabilities but may be overkill for simple string comparisons. Performance overhead might be an issue.
- Method 4: Using os.path.commonprefix. Utilizes a standard library function intended for file paths, but it’s a bit of a workaround and less intuitive for generic strings.
- Bonus Method 5: List Comprehension With Enumerate. Extremely concise, but the complexity of the one-liner can make it less readable. Fits well for those who prefer functional programming styles.