src.valid_anagram

Classes

Solution()

class src.valid_anagram.Solution

isAnagram(s: str, t: str) → bool

Thought process

• Trivially if the words have a different length we return False

• We create two dictionaries to store the frequency of each character in the words

• We compare the dictionaries to see if they are equal

Notes

• time complexity: \(O(n)\)

• space complexity: \(O(n)\)

isAnagramSort(s: str, t: str) → bool

Thought process

• Trivially if the words have a different length we return False

• We sort the words and compare them

Notes

• time complexity: \(O(n^2)\) -> \(O(n\log n)\)

• space complexity: \(O(n)\) -> \(O(1)\)

isAnagramCounter(s: str, t: str) → bool

A bit of a cheat, but we can use the Counter class from the collections module