💡 Problem Formulation: Determining if a given string is an anagram of a palindrome tests one’s ability to assess character arrangements with symmetry. An anagram of a palindrome doesn’t need to form a palindrome itself but rearranging its letters should form one. For instance, the input string ‘arra’ can be rearranged to ‘raar’ or ‘arar’, both of which are palindromes.
Method 1: Use of Counter from Collections
This method involves leveraging the Counter class from Python’s collections module to create a frequency map of the characters in the string. An anagram of a palindrome can have at most one character with an odd count.
Here’s an example:
from collections import Counter
def is_anagram_of_palindrome(s):
freq = Counter(s)
odd_count = sum(1 for char_count in freq.values() if char_count % 2 != 0)
return odd_count <= 1
print(is_anagram_of_palindrome('arra'))
Output: True
This code snippet counts the frequency of each character in the string using the Counter class. The result is ‘True’ for ‘arra’ because it is an anagram of the palindrome ‘raar’.
Method 2: Using defaultdict
With defaultdict from the collections module, this method simplifies counting while still allowing for efficient computation of odd character counts.
Here’s an example:
from collections import defaultdict
def is_anagram_of_palindrome(s):
freq = defaultdict(int)
for char in s:
freq[char] += 1
return sum(count % 2 for count in freq.values()) <= 1
print(is_anagram_of_palindrome('carerac'))
Output: True
This approach constructs a default dictionary to count character occurrences and then checks if there is at most one character with an odd count, which is true for ‘carerac’.
Method 3: Without Collections Module
This method manually counts character occurrences in a dictionary and checks the condition for an anagram of a palindrome without using the collections module.
Here’s an example:
def is_anagram_of_palindrome(s):
freq = {}
for char in s:
freq[char] = freq.get(char, 0) + 1
odd_count = 0
for count in freq.values():
if count % 2 != 0:
odd_count += 1
if odd_count > 1:
return False
return True
print(is_anagram_of_palindrome('deified'))
Output: True
This code manually builds a character count dictionary and counts the number of odd occurrences to verify the palindromic anagram property. ‘deified’ is confirmed as a valid anagram of a palindrome.
Method 4: Optimized Character Count
This method optimizes counting by stopping the character frequency check as soon as more than one odd occurrence is found.
Here’s an example:
def is_anagram_of_palindrome(s):
freq = {}
odd_count = 0
for char in s:
freq[char] = freq.get(char, 0) + 1
if freq[char] % 2 == 0:
odd_count -= 1
else:
odd_count += 1
return odd_count <= 1
print(is_anagram_of_palindrome('radar'))
Output: True
In this code, the frequency map updates and odd count check are performed within the same loop to minimize iterations, therefore improving efficiency slightly. The string ‘radar’ passes the check, indicating that it is an anagram of a palindrome.
Bonus One-Liner Method 5: Using Python Set
This succinct method uses set comprehension and the fact that a palindromic anagram has zero or one odd occurring character.
Here’s an example:
def is_anagram_of_palindrome(s):
return sum(s.count(char) % 2 for char in set(s)) <= 1
print(is_anagram_of_palindrome('tacocat'))
Output: True
Sleek and Pythonic, this one-liner evaluates the frequency of each unique character directly within a generator expression and sums the results. The example ‘tacocat’ showcases a palindrome anagram.
Summary/Discussion
- Method 1: Counter from Collections. Utilizes built-in functionality for counting. Strengths: easy to understand, concise code. Weaknesses: relies on the collections module.
- Method 2: Using defaultdict. Prevents KeyError and simplifies the character counting process. Strengths: no need for extra condition check when incrementing. Weaknesses: Relatively not as straightforward as Counter.
- Method 3: Without Collections Module. Does not depend on external libraries. Strengths: illustrates basic Python dictionary handling. Weaknesses: slightly more verbose and potentially slower.
- Method 4: Optimized Character Count. Provides an immediate exit on finding the second odd count to increase efficiency. Strengths: potentially faster for large strings. Weaknesses: slightly more complex logic.
- Method 5: Python Set One-Liner. Offers a terse, readable solution. Strengths: elegant, very Pythonic. Weaknesses: Can be less efficient on large strings due to repeated counting.