LeetCode 210 - Course Schedule II

Table of Contents

• Problem Statement
• Using Breadth First Search
• Complexity
• Conclusion

Problem Statement

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.

Using Breadth First Search

To determine the order in which courses can be taken, we can use BFS approach for topological sorting. This problem is essentially a graph problem where courses are nodes and prerequisites are directed edges.

1. Graph Construction:

   • Represent courses as nodes and prerequisites as directed edges in the graph.
     
     

2. BFS for Topological Sort:

   • Start with courses that have no prerequisites (in-degree 0).

   • Process these courses, adding their neighbors to the queue when their prerequisites are fulfilled (in-degree becomes 0).

3. Cycle Detection:

   • If not all courses are processed, it indicates a cycle in the graph, making it impossible to complete all courses.

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> adjList(numCourses);
        vector<int> inDegree(numCourses, 0);

        for (auto& prereq : prerequisites) {
            int course = prereq[0];
            int prerequisite = prereq[1];
            adjList[prerequisite].push_back(course);
            inDegree[course]++;
        }
            
        queue<int> q;
        for (int i = 0; i < numCourses; i++) {
            if (inDegree[i] == 0) {
                q.push(i);
            }
        }
            
        vector<int> result;
        while (!q.empty()) {
            int current = q.front();
            q.pop();
            result.push_back(current);

            for (int neighbor : adjList[current]) {
                inDegree[neighbor]--;
                if (inDegree[neighbor] == 0) {
                    q.push(neighbor);
                }
            }
        }
            
        if (result.size() == numCourses) {
            return result;
        } else {
            return {};
        }
    }
};

Complexity

Time Complexity: 𝑂(𝑉 + 𝐸)

• Where 𝑉 is the number of courses and 𝐸 is the number of prerequisites.
• We build the graph in 𝑂(𝐸) and travere the graph in 𝑂(𝑉 + 𝐸).

Space Complexity: 𝑂(𝑉 + 𝐸)

• For the adjacency list and the in-degree tracking.

Conclusion

This approach efficiently computes the course order or detects if that’s impossible to complete all courses due to cyclic dependencies.

Source:

210. Course Schedule II