[LeetCode] 1462. Course Schedule IV (Python)

 

난이도: Medium

문제 설명

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There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course ai first if you want to take course bi.

• For example, the pair [0, 1] indicates that you have to take course 0 before you can take course 1.

Prerequisites can also be indirect. If course a is a prerequisite of course b, and course b is a prerequisite of course c, then course a is a prerequisite of course c.

 

You are also given an array queries where queries[j] = [uj, vj]. For the jth query, you should answer whether course uj is a prerequisite of course vj or not.

 

Return a boolean array answer, where answer[j] is the answer to the jth query.

 

문제 예제

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Example 1:

Input: numCourses = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: The pair [1, 0] indicates that you have to take course 1 before you can take course 0.
Course 0 is not a prerequisite of course 1, but the opposite is true.

Example 2:

Input: numCourses = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites, and each course is independent.

Example 3:

Input: numCourses = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

 

제한 사항

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• 2 <= numCourses <= 100
• 0 <= prerequisites.length <= (numCourses * (numCourses - 1) / 2)
• prerequisites[i].length == 2
• 0 <= ai, bi <= numCourses - 1
• ai != bi
• All the pairs [ai, bi] are unique.
• The prerequisites graph has no cycles.
• 1 <= queries.length <= 104
• 0 <= ui, vi <= numCourses - 1
• ui != vi

 

Solution(솔루션)

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class Solution:
    def checkIfPrerequisite(self, numCourses: int, prerequisites: List[List[int]], queries: List[List[int]]) -> List[bool]:
        graph = [[] for _ in range(numCourses)]
        path = [[] for _ in range(numCourses)]

        for v1, v2 in prerequisites:
            graph[v1].append(v2)
        
        for i in range(numCourses):
            q = deque([i])
            visited = [False] * numCourses
            visited[i] = True

            while q:
                vertex = q.popleft()
                path[i].append(vertex)

                for next_vertex in graph[vertex]:
                    if not visited[next_vertex]:
                        q.append(next_vertex)
                        visited[next_vertex] = True

        answer = []
        for v1, v2 in queries:
            answer.append(v2 in path[v1])
        
        return answer

 

먼저, prerequisites를 기준으로 그래프를 만들어 주었다.

 

그 후 각 vertex 별로 bfs를 통해 갈 수 있는 vertex를 저장해주었다.

 

마지막으로, answer리스트를 만들고 query에 두번째 요소가 저장해두었던 path에 있다고 하면 True를 저장하고 아니면 False를 저장했다.

 

그리고 answer를 return하였다.

 

bfs를 사용하긴 했지만 brute force 느낌으로 푼 것 같다.

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문제: 1462. Course Schedule IV 

 

깃허브: github

algorithmPractice/LeetCode/1558-course-schedule-iv at main · laewonJeong/algorithmPractice