Leetcode Study Day 45

Course Schedule II

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]]
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.

Solution

The solution is similar to the previous question, except this time we add a stack to store the courses we want to take. After we find the courses that need to be taken in the end, we push them into the stack. After finding all the courses and make sure there is no cycle, we pop the stack and store the courses in the vector. The order of the courses is the reverse order of the stack.

class Solution {
public:
    bool iscycle(vector<int> adj[], vector<int> &vis, int id, stack<int> &courseStack) {
        if(vis[id] == 1)
            return true;
        if(vis[id] == 0){
            vis[id] = 1;
            for(int edge : adj[id]){
                if(iscycle(adj, vis, edge, courseStack))
                    return true;
            }
            courseStack.push(id);
        }
        vis[id] = 2;
        return false;
    }

    vector<int> findOrder(int n, vector<vector<int>>& pre) {
        vector<int> adj[n];
        for(auto edge : pre)
            adj[edge[1]].push_back(edge[0]);
        
        vector<int> vis(n, 0);
        stack<int> courseStack;
        for(int i = 0; i < n; i++){
            if(vis[i] == 0){
                if(iscycle(adj, vis, i, courseStack))
                    return {}; // Return empty if cycle is detected
            }
        }

        vector<int> order;
        while(!courseStack.empty()){
            order.push_back(courseStack.top());
            courseStack.pop();
        }
        return order;
    }
};