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hanby-choi
approved these changes
Mar 20, 2024
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| int solution(vector<int>& v, string word, int cnt, string flag) { |
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P3: 문제 상황에 맞게 두 단어의 구성 차이를 계산하는 함수를 잘 정의해 주셨네요! 다만 함수 이름을 지을 때 solution보다는 함수가 어떤 기능을 수행하는지 알 수 있게 지어주시면 더 좋을 것 같습니다. 이 경우 countDiff 같은 이름을 사용할 수 있겠네요!
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| countAlpha(flag, v); | ||
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| for (int i = 1; i < n; i++) { | ||
| string word; | ||
| cin >> word; | ||
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| int diff = solution(v, word, cnt, flag); | ||
| if (diff == 0 || diff == 1 || (diff == 2 && flag.size() == word.size())) { | ||
| cnt++; | ||
| } | ||
| } | ||
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| cout << cnt; |
chaeri93
approved these changes
Mar 22, 2024
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| if (a == 0) { | ||
| if (present.empty()) { | ||
| cout << -1 << "\n"; |
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| if (pq.size() < n) { | ||
| pq.push(num); | ||
| } | ||
| else { | ||
| if (pq.top() < num) { | ||
| pq.pop(); | ||
| pq.push(num); | ||
| } |
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if문 없이 push 해주신다음에 pq의 사이즈가 n을 초과하면 pop을 해주는 로직으로 다시 구현해보면 좋을것같아요,,..!
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인적사항
학번: 2276209
이름: 윤지운
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