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dohyeon2
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Mar 12, 2026
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전반적으로 풀이가 깔끔하네요
3Sum에 부르트포스 접근도 과정이 보여서 흥미롭게 봤습니다.
TC, SC 분석도 함께 남기면 더 공부가 될것같습니다!
3Sum의 새로운 답이 통과 된 이유도 명확해질것같아요.
| return dfs(node.left, min, node.val) && dfs(node.right, node.val, max); | ||
| } | ||
|
|
||
| return dfs(root, Number.NEGATIVE_INFINITY, Number.POSITIVE_INFINITY); |
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Number에 이런 상수가 있는걸 이제 알았네요
배워갑니다~
Cyjin-jani
approved these changes
Mar 13, 2026
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| // const hasArray = result.some( | ||
| // (item) => | ||
| // JSON.stringify([...item].sort((a, b) => a - b)) === | ||
| // JSON.stringify([nums[i], nums[j], nums[k]].sort((a, b) => a - b)), | ||
| // ); |
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오.. 이렇게 중복확인을 할 수도 있겠네요..! 아이디어 얻어갑니다ㅎㅎ
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비록 시간복잡도에서 타임 리밋 걸리긴했지만 저런 방법도 있었습니다
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| threeSum([ | ||
| 12, 5, -12, 4, -11, 11, 2, 7, 2, -5, -14, -3, -3, 3, 2, -10, 9, -15, 2, 14, | ||
| -3, -15, -3, -14, -1, -7, 11, -4, -11, 12, -15, -14, 2, 10, -2, -1, 6, 7, 13, | ||
| -15, -13, 6, -10, -9, -14, 7, -12, 3, -1, 5, 2, 11, 6, 14, 12, -10, 14, 0, -7, | ||
| 11, -10, -7, 4, -1, -12, -13, 13, 1, 9, 3, 1, 3, -5, 6, 9, -4, -2, 5, 14, 12, | ||
| -5, -6, 1, 8, -15, -10, 5, -15, -2, 5, 3, 3, 13, -8, -13, 8, -5, 8, -6, 11, | ||
| -12, 3, 0, -2, -6, -14, 2, 0, 6, 1, -11, 9, 2, -3, -6, 3, 3, -15, -5, -14, 5, | ||
| 13, -4, -4, -10, -10, 11, | ||
| ]); // [[-15,1,14],[-15,2,13],[-15,3,12],[-15,4,11],[-15,5,10],[-15,6,9],[-15,7,8],[-14,0,14],[-14,1,13],[-14,2,12],[-14,3,11],[-14,4,10],[-14,5,9],[-14,6,8],[-14,7,7],[-13,-1,14],[-13,0,13],[-13,1,12],[-13,2,11],[-13,3,10],[-13,4,9],[-13,5,8],[-13,6,7],[-12,-2,14],[-12,-1,13],[-12,0,12],[-12,1,11],[-12,2,10],[-12,3,9],[-12,4,8],[-12,5,7],[-12,6,6],[-11,-3,14],[-11,-2,13],[-11,-1,12],[-11,0,11],[-11,1,10],[-11,2,9],[-11,3,8],[-11,4,7],[-11,5,6],[-10,-4,14],[-10,-3,13],[-10,-2,12],[-10,-1,11],[-10,0,10],[-10,1,9],[-10,2,8],[-10,3,7],[-10,4,6],[-10,5,5],[-9,-5,14],[-9,-4,13],[-9,-3,12],[-9,-2,11],[-9,-1,10],[-9,0,9],[-9,1,8],[-9,2,7],[-9,3,6],[-9,4,5],[-8,-6,14],[-8,-5,13],[-8,-4,12],[-8,-3,11],[-8,-2,10],[-8,-1,9],[-8,0,8],[-8,1,7],[-8,2,6],[-8,3,5],[-8,4,4],[-7,-7,14],[-7,-6,13],[-7,-5,12],[-7,-4,11],[-7,-3,10],[-7,-2,9],[-7,-1,8],[-7,0,7],[-7,1,6],[-7,2,5],[-7,3,4],[-6,-6,12],[-6,-5,11],[-6,-4,10],[-6,-3,9],[-6,-2,8],[-6,-1,7],[-6,0,6],[-6,1,5],[-6,2,4],[-6,3,3],[-5,-5,10],[-5,-4,9],[-5,-3,8... |
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릿코드 테스트 케이스 복사해왔습니다 ㅋㅋ
| @@ -0,0 +1,26 @@ | |||
| function isAnagram(s: string, t: string): boolean { | |||
| if (s.length !== t.length) return false; | |||
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| for (const value in list) { | ||
| if (list[value] > 0) { | ||
| return false; | ||
| } | ||
| } |
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이미 충분히 괜찮은 방법입니다만, 3번째 for문을 없애고 조금 더 빠르게 만들 수 있을 듯 합니다!🤔
이미 s와 t의 길이가 같다보니 아래처럼 t의 문자가 list에 없거나 값이 이미 0인 경우 바로 false를 반환하도록 수정할 수도 있을 것 같아요..!
for (const key of t) {
if (!list[key]) {
return false;
}
list[key] -= 1;
}
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오.. 생각해보지 못했는데, 좋은 방법이네요
한번 돌려봐야겠어요!! 좋은 의견 너무 감사합니다~
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