242. Valid Anagram

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

First solv

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class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        def checker(x:str):
            check = {}
            for i in x:
                if i not in check:
                    check[i] = 1
                else:
                    check[i] += 1
            return check
        check_s, check_t = checker(s), checker(t)
        if check_s == check_t:
            return True
        else:
            return False

It is easy to solve this problem with hash table. Cause need to make hash table exactly same way for s and t, made a checker function. It looks like this {"a":1, "b":2}. The dict which is hash table in python has these features.

1. key never duplicated
2. it is set with value. So no need to be sorted

So with these features, just make 2 dictionary and compared it.

ring my bell

My solution has middle level of runtime and memory efficiency. One solution I saw and ringed my bell is this solution.

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        s_count = Counter(s)
        t_count = Counter(t)

        return s_count == t_count

collections — Container datatypes

Source code: Lib/collections/__init__.py This module implements specialized container datatypes providing alternatives to Python’s general purpose built-in containers, dict, list, set, and tuple.,,…

Python documentation

The Counter in python in built in class for a dict subclass for counting hashable objects. It is exactly what we want for this problem and it is always good to use built in solution in python.