from collections import Counter
class Solution:
def isAnagram(self, s: str, t: str) -> bool:
"""
Determine if string 't' is an anagram of string 's' by comparing character frequencies.
Parameters:
s (str): The first string to compare.
t (str): The second string, which is to be checked as an anagram of 's'.
Returns:
bool: True if 't' is an anagram of 's', otherwise False.
"""
# If lengths are not equal, 't' cannot be an anagram of 's'
if len(s) != len(t):
return False
# Create frequency counters for both strings using Counter (efficient for Unicode as well)
# freq_s counts occurrences of each character in s
freq_s = Counter(s)
# freq_t counts occurrences of each character in t
freq_t = Counter(t)
# Compare the two frequency dictionaries; anagrams have identical counts for every character
return freq_s == freq_t
Summary of Techniques and Approaches:
Hashing with Frequency Counters: By utilizing collections.Counter, the solution creates hash maps to count character occurrences. This approach is efficient (O(n)) and directly applicable for problems that require frequency comparisons or counting elements.
Early Termination by Length Check: The solution first checks if the lengths of the two strings are equal, an effective initial filter to avoid unnecessary computations. This technique is useful in many string-related problems where mismatched sizes immediately disqualify a valid solution.
Adaptability for Unicode: The use of Counter handles Unicode characters seamlessly without additional modifications. This makes the approach broadly applicable when the input might extend beyond simple lowercase English letters.
General Problem-Solving Tip: Look for ways to reduce problem constraints (like eliminating non-candidate comparisons) and use built-in data structures to simplify logic. When faced with problems involving comparisons of element counts, consider using hash maps or frequency counters to deliver clear and efficient solutions.