210. Course Schedule II

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

Example 1:

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

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2|1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• ai != bi
• All the pairs [ai, bi] are distinct.

Approach 1: Optimal BFS

Intuition

• TOPO sort

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 topo
9. Else return {};

Code

class Solution {
public:
    vector<int> findOrder(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 topo;
        } else {
            return {};
        }

    }
};

Complexity Analysis

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

Approach 2: Better

Intuition

Algorithm

Code

Complexity Analysis

• Time Complexity: $O()$
• Space Complexity: $O()$

Approach 3: Optimal

Intuition

Algorithm

Code

Complexity Analysis

• Time Complexity: $O()$
• Space Complexity: $O()$