[printjin-gmailcom] Week 05 Solutions #1378
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soobing
approved these changes
Apr 27, 2025
| def groupAnagrams(self, strs): | ||
| anagrams = defaultdict(list) | ||
| for word in strs: | ||
| count = [0] * 26 |
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오.. 문자 개수 기반으로 그룹화 하는 부분이 인상깊네요.
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시간 복잡도만 보면 카운트 방식이 더 빠를 거라고 생각했는데, 리트코드에서는 문자열을 정렬하는 방법이 더 빠르더라고요. 아마 테스트에 사용된 단어들이 짧아서 그런 것 같습니다만 효율적인 방법 공유해보고자 최종코드로 넣었어요 :)
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