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[혜준] Week2 문제 풀이 #351

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Merged
merged 7 commits into from
Aug 25, 2024
Merged

[혜준] Week2 문제 풀이 #351

merged 7 commits into from
Aug 25, 2024

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hyejjun
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@hyejjun hyejjun commented Aug 20, 2024

  1. Valid Anagram
    시간 복잡도: O(n)
    공간 복잡도: O(n)

  2. Counting Bits
    시간 복잡도: O(n * log n)
    공간 복잡도: O(n)

@hyejjun hyejjun marked this pull request as draft August 20, 2024 03:34
@hyejjun hyejjun added the js label Aug 20, 2024
@DaleSeo
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DaleSeo commented Aug 21, 2024

  • Valid Anagram
    시간 복잡도: O(n)
    공간 복잡도: O(n)
  • Counting Bits
    시간 복잡도: O(n * log n)
    공간 복잡도: O(n)

복잡도 분석은 코드에(도) 주석으로 남겨주시면 리뷰하시는 분들이 더 편할 것 같습니다.

@hyejjun hyejjun marked this pull request as ready for review August 23, 2024 23:13
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@DaleSeo DaleSeo left a comment

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수고하셨습니다!

@DaleSeo DaleSeo requested a review from seona926 August 24, 2024 00:03
console.log(countBits(5));

/*
시간 복잡도: O(n * log n)
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안녕하세요~ 어떤 이유로 시간 복잡도가 n log n 이 나오는지, 여기서 n은 무엇을 가리키는지 설명해주시면 더 좋을 것 같아요! 나중에 실제 코딩 테스트 면접에서도 도움이 될 거라고 생각합니다 ☺️

Comment on lines +7 to +11
let result = [];

for (let i = 0; i <= n; i++) {

let binary = i.toString(2);
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@sounmind sounmind Aug 24, 2024

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Suggested change
let result = [];
for (let i = 0; i <= n; i++) {
let binary = i.toString(2);
�const result = [];
for (let i = 0; i <= n; i++) {
const binary = i.toString(2);

저는 재할당 되지 않을 변수는 const 키워드로 선언하는 편인데요. 이는 코드를 읽는 사람으로 하여금 이 변수는 재할당되지 않을 것임을 예측할 수 있도록 해주고, 나중에 실수로 해당 변수에 다른 값을 할당하게 되는 버그를 사전에 예방할 수 있기 때문이에요.

@DaleSeo DaleSeo merged commit 3e1d624 into DaleStudy:main Aug 25, 2024
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3 participants