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[Jay-Mo-99] Week 5 #861

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@Jay-Mo-99 Jay-Mo-99 commented Jan 6, 2025 •
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답안 제출 문제

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@Jay-Mo-99 Jay-Mo-99 requested a review from jungsiroo January 6, 2025 21:20
@Jay-Mo-99 Jay-Mo-99 self-assigned this Jan 6, 2025
@Jay-Mo-99 Jay-Mo-99 requested a review from a team as a code owner January 6, 2025 21:20
@github-actions github-actions bot added the py label Jan 6, 2025
jungsiroo
jungsiroo reviewed Jan 9, 2025
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상세 설명들을 써두는 점이 좋은 것 같습니다. 시간복잡도와 다양한 테스트 케이스를 고민해보는게 라이브 코딩테스트에서 어렵더라구요,, 이 부분 같이 고민해나가면 좋을 것 같습니다!

encode-and-decode-strings/Jay-Mo-99.py


def encode(self, strs: List[str]) -> str:
answer = '!@#$%123456789'.join(strs)
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만약 strs배열의 구성요소가 ['!@#$%123456789', '!@#$%', ''!@#$%123456789123123'] 이라면 encode가 조금은 꼬이지 않을까 합니다

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네, 구분자(!@#$%123456789) 가 strs의 요소로 나온다면 해당 풀이법은 유효하지 않습니다. 가장 안전한 방법은 strs의 각 요소의 길이를 측정하여 decode때 해당 길이를 기반으로 나눠주는 방법이라고 들었습니다.

group-anagrams/Jay-Mo-99.py
tempDict = defaultdict(list)

for s in strs:
key = tuple(sorted(s))
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정렬을 하지 않고 다르게 풀면 시간복잡도를 줄일 수 있을 것 같은데 어떤 방법이 있을까요?

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