[kimyoung] WEEK 03 Solutions #389
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DaleSeo
approved these changes
Aug 29, 2024
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| for (let i = 1; i <= amount; i++) { | ||
| // only works if coin is less than or equal to the assessing amount | ||
| if (coin <= i) { | ||
| // min comparison | ||
| dp[i] = Math.min(dp[i], dp[i - coin] + 1); | ||
| } | ||
| } |
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i를 coin부터 시작하면 굳이 coin <= i를 체크해줄 필요가 없지 않을까 생각이 들었습니다.
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| for (let i = 1; i <= amount; i++) { | |
| // only works if coin is less than or equal to the assessing amount | |
| if (coin <= i) { | |
| // min comparison | |
| dp[i] = Math.min(dp[i], dp[i - coin] + 1); | |
| } | |
| } | |
| for (let i = coin; i <= amount; i++) { | |
| // min comparison | |
| dp[i] = Math.min(dp[i], dp[i - coin] + 1); | |
| } |
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이제보니 이렇게 최적화가 가능했군요. 앞으로는 코드 마무리 후에 최적화 할수 있는 방법도 고민해보도록 하겠습니다!
| }; | ||
|
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||
| // time - O(2^n) backtracking | ||
| // space - O(1) |
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혹시 curr 배열의 메모리 사용량을 고려하셨나요?
HC-kang
approved these changes
Aug 29, 2024
| for (let i = 2; i <= n; i++) { | ||
| arr[i] = arr[i - 2] + arr[i - 1]; | ||
| } | ||
| return n === 1 ? 1 : arr[n]; |
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안녕하세요 @kim-young 님!
n이 1인지 확인하는 부분은 앞쪽에서 early return 해 주면 가독성이 좀 더 좋아지지 않을까요?
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좋은 리뷰 감사합니다! 이제보니 return statement가 가독성이 많이 떨어지네요. 😂
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