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dalestudy merged 1 commit into from
Aug 30, 2026
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[freemjstudio] WEEK 10 Solutions #2845

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34 changes: 34 additions & 0 deletions course-schedule/freemjstudio.py
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@dalestudy dalestudy Bot Aug 30, 2026

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🏷️ 알고리즘 패턴 분석

course-schedule/freemjstudio.py 
class Solution:
    def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:

        # cycle 을 감지하면 false 를 리턴한다.
        graph = [[] for _ in range(numCourses)]

        # a <- b
        for a, b in prerequisites:
            graph[b].append(a)

        state = [0] * numCourses
        # 0: unvisited 1: visiting 2: done

        def dfs(node):
            if state[node] == 1:
                return False

            if state[node] == 2:
                return True

            state[node] = 1 # visiting

            for next_node in graph[node]:
                if dfs(next_node) == False:
                    return False

            state[node] = 2 # done
            return True

        for course in range(numCourses):
            if dfs(course) == False:
                return False

        return True
  • 패턴: Depth-First Search, Graph
  • 설명: 그래프를 구성하고 DFS로 사이클 여부를 탐지하는 패턴으로, 방문 상태를 추적해 순환 여부를 판단한다. 사이클이 있으면 false, 없으면 true를 반환한다.

📊 시간/공간 복잡도 분석

복잡도
Time O(N + P)
Space O(N + P)

피드백: 그래프를 인접 리스트로 구성하고 각 노드를 DFS하며 사이클 여부를 검사한다. 상태 배열로 방문 중인 노드를 추적해 사이클을 탐지한다.

개선 제안: 현재 구현이 적절해 보입니다.

💡 풀이에 시간/공간 복잡도를 주석으로 남겨보세요!

Original file line number Diff line number Diff line change
@@ -0,0 +1,34 @@
class Solution:
def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:

# cycle 을 감지하면 false 를 리턴한다.
graph = [[] for _ in range(numCourses)]

# a <- b
for a, b in prerequisites:
graph[b].append(a)

state = [0] * numCourses
# 0: unvisited 1: visiting 2: done

def dfs(node):
if state[node] == 1:
return False

if state[node] == 2:
return True

state[node] = 1 # visiting

for next_node in graph[node]:
if dfs(next_node) == False:
return False

state[node] = 2 # done
return True

for course in range(numCourses):
if dfs(course) == False:
return False

return True
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