207. Course Schedule

• Link: https://leetcode.com/problems/course-schedule/description/

Question

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|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.

Approach 1: Optimal

Intuition

• BFS (TOPO sort)
• Return false if a cycle is detected or else true

Algorithm

1. Convert edge list to adjacency list
2. Create vector<int> indegree
3. Loop through the adjacency list & increment indegree for the node everytime that node is reached
4. Declare a queue<int> q
5. Loop through the indegree array
   1. If indegree of the node == 0
      1. Push the node into the queue
6. Declare vector<int> topo
7. While q is not empty
   1. Declare & initalize node with the value of front of the queue
   2. Pop the queue
   3. Push the node into topo
   4. Loop through the neighbors
      1. Decrement neigbor's indegree
      2. If indegree == 0 , push the node into the queue
8. If length of topo sort == numCourses return true
9. Else return false;

Code

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        vector<vector<int>> adj(numCourses, vector<int>());

        for (auto edge: prerequisites) {
            adj[edge[1]].push_back(edge[0]);
        }

        vector<int> indegree(numCourses, 0);

        for (int i = 0 ; i < numCourses ; i++) {
            for (auto neighbor: adj[i]) {
                indegree[neighbor]++;
            }
        }

        queue<int> q;

        for (int i = 0 ; i < numCourses ; i++) {
            if (indegree[i] == 0) {
                q.push(i);
            }
        }

        vector<int> topo;

        while(!q.empty()) {
            int node = q.front();
            q.pop();
            topo.push_back(node);

            for (auto neighbor: adj[node]) {
                indegree[neighbor]--;
                if (indegree[neighbor] == 0) {
                    q.push(neighbor);
                }
            }
        }

        if (topo.size() == numCourses) {
            return true;
        }

        return false;
    }
};

Complexity Analysis

• Time Complexity: $O(V+E)$
• Space Complexity: $O(V+E)$