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obzva
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Jan 30, 2025
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이번 주도 고생 많으셨습니다!
저번 주차 피드백까지 반영하시느라 수고하셨네요 ..
감사합니다!!
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| 정점 V, 간선 E | ||
| Time: O(V + E) | ||
| Space: O(V + E) |
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공간 복잡도 설명을 부탁드려도 될까요? 제 생각엔 O(V)같아서요 :) 채디님 생각이 궁금합니다
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O(V) 만큼의 call stack 사용
O(E) 만큼의 oldToNew 의 value (neighbors) 저장
이라고 생각해 O(V + E) 라고 생각했습니다..!
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| 한번에 풀지 못해 다시 풀어볼 예정입니다. | |||
| Solution: 재귀를 활용한 풀이, memo 를 활용한 시간복잡도 개선 | ||
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| Time: O(m * n) | ||
| Space: O(2 * m * n) |
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일반적으로 Big-O 표기를 할 땐 상수곱을 생략합니다
O(2 * m * n) -> O(m * n)
참고: #445
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2도 어떻게 곱하게 되신건지 여쭤볼 수 있을까요 :)
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| Solution: 재귀를 활용한 풀이, memo 를 활용한 시간복잡도 개선 | |||
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열정이 있으신 것 같아서 팁을 드리자면, tabulation (흔히 bottom-up라고 부르는..)을 이용하면 좀 더 공간 효율적인 풀이가 가능합니다
고민해보시거나, 이번 기수 저번 기수 다른 분들 풀이를 참고해주세요!
| 2) 중복되는 원소를 발견할 경우 해당 원소의 중복이 사라질때까지 left side 의 원소들을 하나씩 제거한다. | ||
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| Time: O(n^2) = O(n) (for iteration) * O(n) 최악의 경우 n만큼의 중복제거 | ||
| Time: O(2n) = O(n) (최악의 경우 n만큼의 중복제거) * O(2) (원소당 최대 2번의 연산) |
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이것도 마찬가지로 O(n)이라고 적어주시는게 맞을 것 같습니다
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