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[okyungjin] WEEK 02 solutions #2683
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| @@ -0,0 +1,32 @@ | ||
| # https://leetcode.com/problems/house-robber/description/ | ||
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| # [요구사항] | ||
| # 문자열 두 개가 주어졌을 때 애너그램이 맞는지 아닌지를 반환하는 문제 | ||
| # 애너그램이란? 하나의 단어나 구에 들어 있는 글자들을 모두, 각각 정확히 한 번씩만 사용해서 순서를 바꾸어 만든 다른 단어나 구를 의미한다. | ||
| # 두 문자열이 애너그램이면 True, 아니면 False 반환 | ||
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| # [접근법] | ||
| # 1. 길이가 다르면 애너그램이 아니므로 바로 False를 반환한다. | ||
| # 2. s의 문자 개수를 센 뒤, t를 순회하며 하나씩 차감한다. | ||
| # 3. 없는 문자이거나 개수가 부족한 문자를 만나면 False를 반환한다. | ||
| # 4. 모든 문자를 정상적으로 처리하면 True를 반환한다. | ||
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| # Time: O(N) | ||
| # Space: O(N) | ||
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| from collections import Counter | ||
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| class Solution: | ||
| def isAnagram(self, s: str, t: str) -> bool: | ||
| if len(s) != len(t): | ||
| return False | ||
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| counter = Counter(s) | ||
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| for char in t: | ||
| if counter[char] >= 1: | ||
| counter[char] -= 1 | ||
| else: | ||
| return False | ||
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| return True |
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🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: Counter를 사용해 s의 문자 빈도를 저장하고 t를 순회하며 차감한다. 문자열 길이가 다르면 빠르게 거르며, 각 문자에 대해 상수 시간 연산을 수행한다.
개선 제안: 현재 구현이 적절해 보입니다.