Course Schedule

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 true if you can finish all courses. Otherwise, return false.

Example 1:

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

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Constraints:

• 1 <= numCourses <= 2000

• 0 <= prerequisites.length <= 5000

• prerequisites[i].length == 2

• 0 <= ai, bi < numCourses

• All the pairs prerequisites[i] are unique.

Solutions

The problem you’re describing is a classic one related to detecting a cycle in a directed graph. Each course can be represented as a node in the graph, and the prerequisite relationship can be represented as a directed edge. If the graph contains a cycle, that means there are some courses that depend on each other in a circular way, so it’s impossible to finish all courses.

Approach - BFS

public class Solution {
    public bool CanFinish(int numCourses, int[][] prerequisites) {
        int[] indegree = new int[numCourses];
        List<int>[] edges = new List<int>[numCourses];
        for (int i = 0; i < numCourses; i++) {
            edges[i] = new List<int>();
        }

        foreach (var prerequisite in prerequisites) {
            indegree[prerequisite[0]]++;
            edges[prerequisite[1]].Add(prerequisite[0]);
        }

        Queue<int> queue = new Queue<int>();
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                queue.Enqueue(i);
            }
        }

        while (queue.Count > 0) {
            int course = queue.Dequeue();
            numCourses--;
            foreach (var nextCourse in edges[course]) {
                if (--indegree[nextCourse] == 0) {
                    queue.Enqueue(nextCourse);
                }
            }
        }

        return numCourses == 0;
    }
}

The time complexity of this algorithm is also O(N + E), where N is the number of courses and E is the number of prerequisites.

The space complexity of the algorithm is O(N + E), where N is the number of courses and E is the number of prerequisites. This is because we are storing the edges and indegrees of the graph, which in total can be up to N + E. Specifically:

• The indegree array takes O(N) space.

• The edges list takes O(E) space.

• The queue can hold all nodes in the worst case, so it also takes O(N) space.

So, the total space complexity is O(N + E).