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] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

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 <= a_i, b_i < numCourses
a_i != b_i
All the pairs [a_i, b_i] are distinct.

Solution:

Construct adjacency list graph and inDegrees array. Then run Kahn’s algorithm.

Kahn’s algorithm

Create two arrays noIncoming and doable both empty initially
Put all courses which have no prerequisites in noIncoming
While noIncoming is not empty
- Pop a node from noIncoming and insert in doable
- Traverse through adjacency list of this node and for each destination node in this adjacency list:
  - Decrement inDegree of this destination node. This is equivalent to breaking an edge
  - if InDegree is 0 then insert it into noIncoming

 
class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) 
    {
        unordered_map<int,vector<int>> G;
        unordered_map<int,int> inDegrees;
        vector<int> noIncoming, doable;
        
        for (auto p: prerequisites)
        {
            G[p[1]].push_back(p[0]);
            inDegrees[p[0]]++;
        }
        
        for (int i=0; i<numCourses; i++)
            if (inDegrees[i]==0)
                noIncoming.push_back(i);
        
        while (noIncoming.size()>0)
        {
            int curr = noIncoming.back();
            noIncoming.pop_back();
            doable.push_back(curr);
            for (int nbd: G[curr])
            {
                inDegrees[nbd]--;
                if (inDegrees[nbd]==0)
                    noIncoming.push_back(nbd);
            }
        }
        if (doable.size() < numCourses) return {};
        return doable;
    }
};

DFS

Here we will need to maintain state of node during DFS. We do it using three colors: White(0) for not visited, Black(1) for visited and completed and Grey(2) for currently visiting and inside recursion.

 
// Colors: 0-Not visited, 1-visited and done, 2-visiting and recursion going on
class Solution {
    bool possible = true;

public:
    void dfs(unordered_map<int,vector<int>> &G, unordered_map<int,int> &colors, int curr, vector<int> &doable)
    {
        if (!this->possible) return;
        colors[curr] = 2;
        for (int nbd: G[curr])
        {
            if (colors[nbd]==0)
                dfs(G,colors,nbd,doable);
            else if (colors[nbd]==2)
                this->possible = false;
        }
        colors[curr] = 1;
        doable.push_back(curr);
    }

    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) 
    {
        unordered_map<int,vector<int>> G;
        unordered_map<int,int> colors;
        vector<int> doable;
        
        for (auto p: prerequisites)
            G[p[1]].push_back(p[0]);
        for (int i=0; i<numCourses; i++)
            colors[i] = 0;

        for (int curr=0; curr<numCourses; curr++)
        {
            if (colors[curr]==0)
                dfs(G, colors, curr, doable);
        }
        
        if (!this->possible) return {};
        reverse(doable.begin(), doable.end());
        return doable;
    }
};