[환미니니] Week4 문제풀이 #425
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[환미니니] Week4 문제풀이 #425
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naringst
reviewed
Sep 6, 2024
| nums.sort((a,b) => a - b); | ||
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| for (let i = 0 ; i <= nums.length; i++) { | ||
| if (i !== nums[i]) return i |
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숫자가 0부터 시작하니까 이렇게 비교해서 풀어도 좋겠네요! 배워갑니다 :)
| * @param {number[]} nums | ||
| * @return {number} | ||
| */ | ||
| var missingNumber = function(nums) { |
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js에서 sort는 내부적으로 추가적인 메모리를 사용해서 공간복잡도가 O(1)은 아닐 것 같습니다!
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엇 그렇네요 정렬할 때 공간복잡도가 On 사용됩니다!
감사합니다 : )
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| while (leftIdx <= rightIdx) { | ||
| if (strs[leftIdx] !== strs[rightIdx]) return false | ||
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오 양쪽에서 인덱스를 증가시키거나 감소시키면서 하나씩 비교하니 풀이가 굉장히 깔끔하네요!
| let maxSequenceLength = -Infinity | ||
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| const setNums = [...new Set(nums)].toSorted((a,b) => a - b) |
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엇 그런데 문제 조건에서 O(n)에 동작하는 코드를 작성해야 한다고 했는데, 해당 코드는 시간복잡도가 nlogn이 되는 것 같습니다!
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앗 그냥 생각나는데로 풀어버렸네요 ㅜ
정렬을 안쓰고 풀어봐야겠어요 ~
DaleSeo
approved these changes
Sep 7, 2024
word-search/hwanmini.js
Outdated
| // m은 board의 행 수, n은 board의 열 수, 4(상하좌우), l(word) | ||
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| // 시간복잡도: O(m * n * 4L) | ||
| // 공간복잡도: O(m * n + L) |
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