242. Valid Anagram

Description

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

Example 1:

• Input: s = “anagram”, t = “nagaram”
• Output: true

Example 2:

• Input: s = “rat”, t = “car”
• Output: false

Constraints:

• 1 <= s.length, t.length <= 5 * 10⁴
• s and t consist of lowercase English letters.

Follow up: What if the inputs contain Unicode characters? How would you adapt your solution to such a case?

Submitted Code

class Solution(object):
    def isAnagram(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        if len(s) != len(t):
            return False
        
        s_letters = {}
        t_letters = {}
        
        for i in range(len(s)):
            if s[i] in s_letters:
                s_letters[s[i]] += 1
            else:
                s_letters[s[i]] = 1

            if t[i] in t_letters:
                t_letters[t[i]] += 1
            else:
                t_letters[t[i]] = 1

        return s_letters == t_letters

Runtime: 15 ms | Beats 83.33%
Memory: 14.39 MB | Beats 14.64%

딕셔너리로 각 글자 수를 세고 비교하는 방법을 사용했다.

Other Solutions

1st

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:  
        if len(s) != len(t):
            return False

        counter = {}

        for char in s:    # s로 딕셔너리 생성
            counter[char] = counter.get(char, 0) + 1

        for char in t:    # t를 체크하면서 값을 하나씩 감소
            if char not in counter or counter[char] == 0:
                return False
            counter[char] -= 1

        return True

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑘) ← 고유 문자 수(알파벳 소문자로 제한될 경우 26)

딕셔너리를 하나만 만드는 방법

2nd

from collections import Counter

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

파이썬에 문자열의 글자수를 딕셔너리처럼 세는 기능이 있다.