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# 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:**

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

**Example 2:**

<pre><code><strong>Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
</strong><strong>Output: false
</strong><strong>Explanation: There are a total of 2 courses to take. 
</strong>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.
</code></pre>

&#x20;

**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

```csharp

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).