LeetCode - Course Schedule II

문제 설명

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 bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

입출력 예시

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

제약사항

1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= ai, bi < numCourses
ai != bi
All the pairs [ai, bi] are distinct.

Solution

[전략]
1. key를 기준으로 다음 course, 이전 course 집합을 set으로 관리
2. 이전 course가 없는 요소를 시작점으로 탐색

    HashMap<Integer, Set<Integer>> fromHash = new HashMap<>(); // key를 기준으로 다음 course
        HashMap<Integer, Set<Integer>> toHash = new HashMap<>(); // ket를 기준으로 이전 course

        for(int i = 0; i < prerequisites.length; i++) {
            Set<Integer> fromSet = toHash.get(prerequisites[i][0]);
            if (fromSet == null) {
                fromSet = new HashSet<>();
                toHash.put(prerequisites[i][0], fromSet);
            }
            fromSet.add(prerequisites[i][1]);
            
            Set<Integer> toSet = fromHash.get(prerequisites[i][1]);
            if (toSet == null) {
                toSet = new HashSet<>();
                fromHash.put(prerequisites[i][1], toSet);
            }
            toSet.add(prerequisites[i][0]);
        }

        Queue<Integer> courseCandidates = new LinkedList();
        for (int i = 0; i < numCourses; i++) {
            if (!toHash.containsKey(i)) {
                courseCandidates.add(i);
            }
        }

        if (courseCandidates.size() == 0) { // 시작점이 없이 순환됨
            return new int[0];
        }
        
        ArrayList<Integer> courseOrder = new ArrayList<>();

        Set<Integer> orderedSet = new HashSet<>();
        while (!courseCandidates.isEmpty()) {
            int orderCandidate = courseCandidates.poll();
            if (orderedSet.contains(orderCandidate)) {
                continue; // 이미 등록됨
            }
            if (toHash.containsKey(orderCandidate)) {
                //skip. 아직 선수과목이 남음
                continue;
            }
            courseOrder.add(orderCandidate);
            orderedSet.add(orderCandidate);

            if (fromHash.containsKey(orderCandidate)) {
                for (int nextVal : fromHash.get(orderCandidate)) {
                    Set<Integer> fromSet = toHash.get(nextVal);
                    if (fromSet == null || !fromSet.contains(orderCandidate)) {
                        // there is cycle;
                        return new int[0];
                    }
                    if (fromSet.size() == 1) {
                        toHash.remove(nextVal);
                    } else {
                        fromSet.remove(orderCandidate);
                    }
                    courseCandidates.add(nextVal);
                }
            }
        }
        if (courseOrder.size() < numCourses) {
            return new int[0];
        }
        return courseOrder.stream().mapToInt(Integer::intValue).toArray();