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KwonNayeon
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Apr 11, 2025
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| """ | ||
| - Algorithm | ||
| - Sort and compares with three pointers: target, left(l), right(r) | ||
| - Time Complexity: O(n^2), n = len(nums) | ||
| - sort : O(nlogn) | ||
| - nested two loops : O(n^2) | ||
| - O(nlogn + n^2) => O(n^2) | ||
| - Space Complexity: O(n^2) if result included. | ||
| - result size : result.append() called in n^2 times (nested two loops) | ||
| """ |
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@ayosecu 님 안녕하세요! 저도 투포인터를 활용하여 문제를 풀었습니다. 개인적으로 알고리즘 문제를 풀 때 tc, sc를 생각해내는 게 항상 어려운데, 이 부분을 잘 정리해주셔서 복습하는 데 도움이 많이 됐습니다!
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이 문제는 반복문으로도 풀어보시면 좋을 것 같습니다! 그리고 항상 테스트 케이스를 기록하시는 모습이 인상 깊었고, 그 부분에서 많이 배웠습니다!
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| def productExceptSelfN(self, nums: List[int]) -> List[int]: | ||
| n = len(nums) | ||
| prefix, suffix, result = [0] * n, [0] * n, [0] * n | ||
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| # Calculate prefix and suffix production | ||
| prefix[0], suffix[-1] = nums[0], nums[-1] | ||
| for i in range(1, n - 1): | ||
| prefix[i] = prefix[i - 1] * nums[i] | ||
| j = n - i - 1 | ||
| suffix[j] = suffix[j + 1] * nums[j] | ||
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| # Update the result | ||
| result[0], result[-1] = suffix[1], prefix[-2] | ||
| for i in range(1, n - 1): | ||
| result[i] = prefix[i - 1] * suffix[i + 1] | ||
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| return result |
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저는 두 번째 풀이 방식으로 문제를 풀었는데, 이 풀이는 제가 생각하지 못했던 방법이라 보면서 공부가 되었습니다! 2주차도 고생하셨습니다 👍
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리뷰 너무 감사합니다! 이번주도 고생하셨고, 다음주도 화이팅입니다! 😄
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