[JANGSEYEONG] WEEK 01 solutions #1162
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| // 배열 전체 돌면서 조건에 맞게 return 시키기 - O(n) | ||
| return nums.map((x, i) => { | ||
| // 0이 2개 이상이라면 무조건 0 | ||
| if (zeros.size > 1) { |
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(이미 빠른 코드이지만) 전체곱을 구하는 reduce에서 0의 개수를 미리 카운트 해놓으면 0이 2개 이상인경우
map을 돌지 않고 함수를 종료 할 수 있어보입니다!
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아 그러네요 !! 0개일 땐 위에서 미리 return 시켜도 무방한 코드인데 map 돌면서 조건 하나하나 분기하면서 생각하다보니 넣어버린 것 같습니다 ㅎㅎ.. 리뷰 감사합니다!
| @@ -0,0 +1,42 @@ | |||
| /* | |||
| 시간복잡도: O(n²) | |||
| - nums.indexOf()는 배열 전체를 순회하는 O(n) 작업 | |||
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O(n^2) 복잡도로는 시간초과가 나는지 궁금합니다..!
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리트코드에서는 시간 초과가 안납니다!
O(n²)일 떄: 75ms, O(n)일 때: 2ms 로 차이가 굉장히 많이 나는건 확인했습니다 ㅎ_ㅎ
문제 하단에 "Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?" 라고 되어있어서 더 줄여봤습니다!
| var containsDuplicate = function (nums) { | ||
| // Set으로 만들었을 때, 기존 배열과 사이즈가 다르면 중복이 제거된거임 | ||
| const numsSet = new Set(nums); | ||
| return nums.length !== numsSet.size; |
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시간, 공간복잡도를 모두 잡은 코드네요 👍👍
YoungSeok-Choi
approved these changes
Apr 4, 2025
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