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[장희직] - 괄호 제거, 도시 건설, 싸지방에 간 준하, 타일 채우기 #262

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merged 1 commit into from
May 21, 2024

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@jhg3410 jhg3410 commented May 20, 2024

📌 from issue #261 📌

📋문제 목록📋

괄호 제거: ✅
도시 건설�: ✅
싸지방에 간 준하�: ✅
타일 채우기: ⛔️

@jhg3410 jhg3410 added the 희직 label May 20, 2024
@jhg3410 jhg3410 self-assigned this May 20, 2024
Comment on lines +19 to +24
if char == ')':
cnt -= 1
if cnt == 0:
return i + 1 + idx
if char == '(':
cnt += 1
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스택 같은 자료구조를 따로 쓰는 게 아니라 카운트로만 이렇게 할 수 있군요

people = []
seat = []
possible_index = []
answer = [0 for _ in range(n + 1)]
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애초에 컴퓨터를 일단 n개로 가정하는 것도 좋네요

dp[x] += 2 * dp[i]
dp[x] += 2

print(dp[n])
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공유해주신 참고자료가 이해가 잘 됐습니다👍

Comment on lines +18 to +22
while parentA != parents[parentA]:
parentA = parents[parentA]

while parentB != parents[parentB]:
parentB = parents[parentB]
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바로 여기서 부모를 찾아도 괜찮군요!

heapq.heappush(seat, [end, idx])
answer[idx] += 1

count_seat = answer.index(0)
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오... index 좋네요

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고생하셨습니다!!


count_seat = answer.index(0)
print(count_seat)
print(*answer[:count_seat])
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헉 이거 포인터처럼 쓰는건가요?...

  • 이거요!

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* <- 이걸 말씀하신 것 같은데
저건 언패킹 연산자라고 배열을 풀어주는 역할을 합니다!
참고로 코틀린에도 스프레드 연산자라고 불리는 비슷한 친구가 있어요!

val array = arrayOf(1,2,3,4)
val list = listOf(*array)
println(list)

>> [1, 2, 3, 4]

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오 대박... 감사합니다.

dp[x] += (dp[x - 2] * dp[2])
for i in range(2, x - 2, 2):
dp[x] += 2 * dp[i]
dp[x] += 2
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요 부분 덕분에 이해했습니다!! :) 최고

@jhg3410 jhg3410 merged commit aadb6c3 into main May 21, 2024
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4 participants