[선재] WEEK09 Solution #515
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[선재] WEEK09 Solution #515
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wogha95
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Oct 12, 2024
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| if (answer < nums[nums.length - 1]) return answer; | ||
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| for (let i = nums.length - 1; i >= 0; i--) { | ||
| if (answer < nums[i]) return answer; | ||
| answer = nums[i]; | ||
| } |
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answer < nums[nums.length - 1] 조건문이 for문 안에 있는 if문에 포함되어 제거 가능해보입니다!
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오호?! 그러네요 중복되는 로직이었군요..!
감사합니다.
| * | ||
| * n = length of height | ||
| * m = length of height[i] | ||
| * time complexity: O(n*m * 4^n*m) |
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상하좌우에 대한 4방향 접근으로 인해 4^n*m으로 계산하신걸까요?
visited 를 사용해서 memoization 하니 각 방향이n*m만큼 탐색일까 하는 의문이 듭니다!
혹시 가능하시다면 시간 복잡도 계산 과정 설명 부탁드립니다!
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안녕하세요. @wogha95 님! 리뷰 코멘트가 늦어진점 죄송합니다🥲
코멘트에 대한 리뷰는 다음과 같아요!
pacific이며 atlantic인 경우에만 memization이 되도록 구현했어요.
그래서 각 인덱스마다 접근시 memization이 안되는 경우가 존재하며 최악의 상황의 시간복잡도는 O(nm * 4^nm) 으로 계산했어요.
추가로 visited는 전역에서 사용하는 변수이지만 해당 bfs 스코프에서 다룰수 있게 값을 설정하여 매번 4방향을 탐색하도록 동작하도록 구현했어요.
그래서 제가 계산한 시간복잡도는 O(nm * 4^nm) 인데 이해가 됐을까요?
@Sunjae95 님, Week 설정 좀 까먹지 말고 좀 부탁드립니다 🥲
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