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[2023-10-02] wooyeol #296 #314
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zsmalla
approved these changes
Oct 8, 2023
| def solution2(prices): | ||
| # answer를 max 값으로 초기화 | ||
| # ex) [1,2,3,2,3] -> [4,3,2,1,0] | ||
| answer = [v for v in range(len(prices)-1, -1, -1)] |
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이렇게 먼저 역으로 한 인덱스에서 가질 수 있는 최대값을 정의해주는 방법이 훨씬 와닿았던 것 같습니다!
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저도 이 부분을 스택에 남은 원소들에 대해 처리해줬었는데 이렇게 구현하는게 훨씬 간결하고 가독성이 좋은 것 같아요!! 고생하셨습니다 우열님 👍👍
ksumini
approved these changes
Oct 9, 2023
| def solution2(prices): | ||
| # answer를 max 값으로 초기화 | ||
| # ex) [1,2,3,2,3] -> [4,3,2,1,0] | ||
| answer = [v for v in range(len(prices)-1, -1, -1)] |
There was a problem hiding this comment.
저도 이 부분을 스택에 남은 원소들에 대해 처리해줬었는데 이렇게 구현하는게 훨씬 간결하고 가독성이 좋은 것 같아요!! 고생하셨습니다 우열님 👍👍
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PR Summary
주식가격
https://school.programmers.co.kr/learn/courses/30/lessons/42584?language=python3
풀이시간
11:30 ~ 11:50 (20분)
문제 조건
1 <= prices의 각 가격 <= 10,000
2 <= prices의 길이 <= 100,000
시간 복잡도 :
O(N(N+1)/2)
접근법
무슨 알고리즘으로 풀이 할 수 있을까? -> 큐 와 스택
문제를 잘못 읽어 10,000개의 데이터 인줄 알고 시간 복잡도를 생각하지않고 풀이했는데
다시보니 100,000개인 것을 확인하였습니다.
그럼 이 풀이로는 풀이가 되지 않아야하는 의아함이 들어 다른 풀이를 참조하였습니다.
풀이 참고 : https://velog.io/@soo5717/프로그래머스-주식가격-Python
해당 풀이를 참고했을 때 2배 정도 빠른 효율성 테스트 결과를 확인 할 수 있었습니다.