Topological sort

Problem Statement

Given a Directed Acyclic Graph (DAG) with V vertices and E edges, Find any Topological Sorting of that Graph.

Example 1:

Input:

Output:
1
Explanation:
The output 1 denotes that the order is
valid. So, if you have, implemented
your function correctly, then output
would be 1 for all test cases.
One possible Topological order for the
graph is 3, 2, 1, 0.

Example 2:

Input:

Output:
1
Explanation:
The output 1 denotes that the order is
valid. So, if you have, implemented
your function correctly, then output
would be 1 for all test cases.
One possible Topological order for the
graph is 5, 4, 2, 1, 3, 0.

Your Task: You don't need to read input or print anything. Your task is to complete the function topoSort() which takes the integer V denoting the number of vertices and adjacency list as input parameters and returns an array consisting of the vertices in Topological order. As there are multiple Topological orders possible, you may return any of them. If your returned topo sort is correct then the console output will be 1 else 0.

Expected Time Complexity: O(V + E). Expected Auxiliary Space: O(V).

Constraints: 2 ≤ V ≤ 104 1 ≤ E ≤ (N*(N-1))/2

Intuition

Approach 1:

DFS traversal, till end 
add end elements to stack, So on, Goes second last etc, till start

Approach 2:
Kahn Alg

Track indegree of each node, Start from each node with 0 indegree the start
points and push them, remove indeg of all its chlidren, If child indeg becomes
zero do for them

Video Links

Approach 1:

C++

class Solution
{
public:
    void dfs(int node, vector<bool> &visited, stack<int> &s, vector<int> adj[]){
        visited[node] = true;
        
        for(auto &it: adj[node]){
            if(!visited[it]){
                dfs(it, visited, s, adj);
            }
        }
        s.push(node);
    }

	vector<int> topoSort(int V, vector<int> adj[]) 
	{
	    vector<bool> visited(V,false);
	    stack<int> s;
	    vector<int> ans;
	    
	    for(int i=0; i<V; i++)
	        if(!visited[i])
	            dfs(i,visited,s,adj);
	    
	    while(!s.empty()){
	        ans.push_back(s.top());
	        s.pop();
	    }
	    return ans;
	}
};

Approach 2:

C++

class Solution
{
public:
    vector<int> topoSort(int V, vector<int> adj[]) {
        vector<int> indeg(V,0);
        queue<int> q;
        // Put all the indegree values by using this loop
        for(int i=0; i<V; i++)
            for(auto &it: adj[i])
                indeg[it]++;
                
        // Start with nodes not having any degree as they are coming first
        for(int i=0; i<V; i++)
            if(indeg[i] == 0)
                q.push(i);
                
        vector<int> ans;
        // Take one by one element and remove the degree of the children        
        while(!q.empty()){
            int temp = q.front();
            
            for(auto &it: adj[temp]){
                indeg[it]--;
                if(indeg[it] == 0)
                    q.push(it);
            }
            ans.push_back(temp);
            q.pop();
        }
        return ans;
    }
};

Approach 3:

C++

Approach 4:

C++