💡 Problem Formulation: Given two strings, how can we determine if the frequency of any character in one string is a factor or a multiple of the frequency of the same character in the other string? For instance, if the character ‘a’ appears 2 times in the first string and 4 times in the second, we conclude that its frequency in the second is a multiple of its frequency in the first.
Method 1: Using a Custom Function to Check Multiplicity
This method involves writing a custom function that takes two strings as arguments and checks for each character if the frequency in one string is a multiple or factor of its frequency in the other string. The method returns a dictionary with characters as keys and Boolean values indicating whether the multiplicity condition holds.
Here’s an example:
def check_multiplicity(str1, str2):
def is_multiple_or_factor(n1, n2):
return n1 % n2 == 0 or n2 % n1 == 0
result = {}
for char in set(str1 + str2):
f1, f2 = str1.count(char), str2.count(char)
if f1 and f2:
result[char] = is_multiple_or_factor(f1, f2)
else:
result[char] = False
return result
# Example usage:
result = check_multiplicity('aabbcc', 'abcabc')
print(result)Output:
{
'a': True,
'b': True,
'c': True
}This function, check_multiplicity(), iterates through all unique characters found in either string, checking and storing whether the frequency of each character in one string is a factor or multiple of its appearance in the other string. The result is a clear, easy-to-interpret dictionary mapping characters to Boolean values.
Method 2: Using a Dictionary Comprehension
Dictionary comprehensions provide a concise way to create dictionaries. Here, we can build a similar dictionary to the former method that indicates for each character if the multiplicity condition is met.
Here’s an example:
str1 = 'aabbcc'
str2 = 'abcabc'
multiplicity_checks = {
char: (str1.count(char) % str2.count(char) == 0 or str2.count(char) % str1.count(char) == 0)
for char in set(str1 + str2) if str1.count(char) and str2.count(char)
}
print(multiplicity_checks)Output:
{
'a': True,
'b': True,
'c': True
}Using a dictionary comprehension we can condense the logic implemented in Method 1. However, the str1.count(char) and str2.count(char) calls are not efficient for long strings, as they have to iterate through the entire strings for each character.
Method 3: Using Counter from Collections
The Counter class from the collections module is an alternative for calculating the frequency of each character. This method is more efficient than repeated calls to the str.count() method, especially for longer strings.
Here’s an example:
from collections import Counter
def check_factorial_relationship(str1, str2):
counter1 = Counter(str1)
counter2 = Counter(str2)
return {
char: ((counter1[char] % counter2[char] == 0 or counter2[char] % counter1[char] == 0) if counter2[char] else False)
for char in counter1
}
result = check_factorial_relationship('aabbcc', 'abcabc')
print(result)Output:
{
'a': True,
'b': True,
'c': True
}The check_factorial_relationship() function first calculates the frequency of characters using the Counter class, which improves efficiency. It then constructs a dictionary that reflects the multiplicity condition for frequencies, as seen in previous methods.
Method 4: Using map and lambda Functions
This method leverages the map function in combination with a lambda to check each character’s frequency condition in a list comprehension. The result is again a dictionary with the desired Boolean values.
Here’s an example:
str1, str2 = 'aabbcc', 'abcabc'
is_multiple_or_factor = lambda f1, f2: f1 % f2 == 0 or f2 % f1 == 0
result = {char: is_multiple_or_factor(str1.count(char), str2.count(char)) for char in set(str1) & set(str2)}
print(result)Output:
{
'a': True,
'b': True,
'c': True
}With the lambda function, is_multiple_or_factor(), we maintain readability while simplifying the operation to check condition across the intersection of unique characters in both strings. The use of set() operations ensure we only process unique characters that exist in both strings.
Bonus One-Liner Method 5: Leveraging generator expressions and all()
A one-liner approach can be taken using generator expressions with the all() function spanning the complete set of unique characters to determine if the condition is universally true across common characters.
Here’s an example:
str1, str2 = 'aabbcc', 'abcabc' result = all((str1.count(char) % str2.count(char) == 0 or str2.count(char) % str1.count(char) == 0) for char in set(str1) & set(str2)) print(result)
Output:
True
This one-liner checks the conditional factor and multiple relationship using all() to confirm if the condition holds true for every character present in both strings. It’s compact and elegant, although it prioritizes brevity over clarity and may be less readable to beginner Pythonistas.
Summary/Discussion
- Method 1: Custom Function with Multiplicity Check. Straightforward and easy to understand. Not as efficient for long strings due to multiple
str.count()calls. - Method 2: Dictionary Comprehension. Concise, one-liner approach inside a comprehension. Efficiency still suffers for long strings due to repeated calls to
str.count(). - Method 3: Using Counter from Collections. Highly efficient and suitable for long strings. Requires an additional import but enhances performance significantly.
- Method 4: Using map and lambda Functions. Provides a compact solution that uses set operations to focus on common characters. Readability may be slightly reduced for those unfamiliar with lambda functions.
- Bonus Method 5: One-Liner with all(). Most succinct solution. May lack readability but offers a quick way to check the universal condition across characters.