1136. Parallel Courses

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Problem

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given an array relations where relations[i] = [prevCourse_i, nextCourse_i], representing a prerequisite relationship between course prevCourse_i and course nextCourse_i: course prevCourse_i has to be taken before course nextCourse_i.

In one semester, you can take any number of courses as long as you have taken all the prerequisites in the previous semester for the courses you are taking.

Return the minimum number of semesters needed to take all courses. If there is no way to take all the courses, return -1.

Example 1:

img

Input: n = 3, relations = [[1,3],[2,3]]
Output: 2
Explanation: The figure above represents the given graph.
In the first semester, you can take courses 1 and 2.
In the second semester, you can take course 3.

Example 2:

img

Input: n = 3, relations = [[1,2],[2,3],[3,1]]
Output: -1
Explanation: No course can be studied because they are prerequisites of each other.

Constraints:

• 1 <= n <= 5000
• 1 <= relations.length <= 5000
• relations[i].length == 2
• 1 <= prevCourse_i, nextCourse_i <= n
• prevCourse_i != nextCourse_i
• All the pairs [prevCourse_i, nextCourse_i] are unique.

Code

210. Course Schedule II

class Solution {
    public int minimumSemesters(int n, int[][] relations) {
        int[] indegree = new int[n + 1];
        for(int[] relation : relations) {
            indegree[relation[1]]++;
        }

        Queue<Integer> queue = new LinkedList<>();

        for(int i = 1; i < indegree.length; i++) {
            if(indegree[i] == 0) {
                queue.offer(i);
                n--;
            }
        }

        int res = 0;
        while(!queue.isEmpty()) {
            int size = queue.size();

            for(int i = 0; i < size; i++) {
                int curr = queue.poll();

                for(int[] relation : relations){
                    if(relation[0] == curr){
                        indegree[relation[1]]--;

                        if(indegree[relation[1]] == 0){
                            queue.offer(relation[1]);
                            n--;
                        }
                    }
                }
            }

            res++;
        }

        if(n == 0) return res;

        return -1;
    }
}