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Data Structure & Algorithm

Check if we can visit all other nodes from any node in given Directed Graph

By Thapelo Manthata
0
3

Given N nodes, where each of them is numbered from 0 to N ā€“ 1, and array edges, where there is a directed edge from edges[i][0] to edges[i][1], the task is to find whether we can travel from any node to all other nodes or not.

Examples:

Input: N = 2, edges[] = {{0, 1}, {1, 0}};
Output: True
Explanation: We can go to node 0 from 1 and node 1 from 0

Input: N = 3, edges[] = {{1, 0}};
Output: False

An approach using Kosarajuā€™s algorithm:

An idea of solving this problem is to think in terms of finding strongly connected component (SCC) for directed graph, We know that a directed graph is strongly connected if there is a path between all pairs of vertices.Ā 

So if there is only a single SCC then only we can visit all the nodes from any other node. The number of SCC can be found using Kosarajuā€™s algorithm.

Follow the steps below to implement the above idea:

  • Create adjacency list adj1 for storing the graph.
  • Do DFS in random order of vertices and store the visited vertices in a stack stk while backtracking.
  • Reverse the direction of all edges of the adj1 graph and store the newly created graph in adj2.
  • Do dfs2 in order of stack and keep count of the strongly connected components in variable scc.
  • Check the count of scc:
    • If the count of scc is greater than 1, return false.
    • Otherwise, return true.

Below is the implementation of the above approach:

C++




// C++ code to implement above approach
Ā 
#include <bits/stdc++.h>
using namespace std;
Ā 
// Function to find components
void randomOrderDfs(int i, vector<vector<int> >& adj,
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā vector<bool>& visited, stack<int>& stk)
{
Ā Ā Ā Ā visited[i] = true;
Ā 
Ā Ā Ā Ā for (auto child : adj[i]) {
Ā Ā Ā Ā Ā Ā Ā Ā if (visited[child] == false) {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(child, adj, visited, stk);
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā }
Ā 
Ā Ā Ā Ā stk.push(i);
}
Ā 
// Function to traverse in the reversed graph
void dfs2(int i, vector<vector<int> >& adj,
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā vector<bool>& visited)
{
Ā Ā Ā Ā visited[i] = true;
Ā 
Ā Ā Ā Ā for (auto child : adj[i]) {
Ā Ā Ā Ā Ā Ā Ā Ā if (visited[child] == false) {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(child, adj, visited);
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā }
}
Ā 
// Function to check if it is possible
// to reach all other nodes from any node
bool isTourPossible(int n, vector<vector<int> >& roads)
{
Ā Ā Ā Ā // adj1 stores adjacency matrix of
Ā Ā Ā Ā // original graph adj2 stores
Ā Ā Ā Ā // adjacency matrix of original graph
Ā Ā Ā Ā // by reversing direction of all edges
Ā Ā Ā Ā vector<vector<int> > adj1(n), adj2(n);
Ā 
Ā Ā Ā Ā // Create graph
Ā Ā Ā Ā for (auto i : roads) {
Ā Ā Ā Ā Ā Ā Ā Ā adj1[i[0]].push_back(i[1]);
Ā Ā Ā Ā }
Ā 
Ā Ā Ā Ā vector<bool> visited1(n, false), visited2(n, false);
Ā Ā Ā Ā stack<int> stk;
Ā 
Ā Ā Ā Ā // Random dfs and maintain stack
Ā Ā Ā Ā // at backtracking (endpoint)
Ā Ā Ā Ā for (int i = 0; i < n; i++) {
Ā Ā Ā Ā Ā Ā Ā Ā if (visited1[i] == false) {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(i, adj1, visited1, stk);
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā }
Ā 
Ā Ā Ā Ā // Reverse all the edges
Ā Ā Ā Ā for (int i = 0; i < n; i++) {
Ā Ā Ā Ā Ā Ā Ā Ā for (auto child : adj1[i]) {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2[child].push_back(i);
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā }
Ā 
Ā Ā Ā Ā // scc for counting the number of
Ā Ā Ā Ā // strongly connected component
Ā Ā Ā Ā int scc = 0;
Ā 
Ā Ā Ā Ā // Make second dfs at order of stk
Ā Ā Ā Ā while (stk.size() > 0) {
Ā Ā Ā Ā Ā Ā Ā Ā int node = stk.top();
Ā Ā Ā Ā Ā Ā Ā Ā if (visited2[node]) {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop();
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā Ā Ā Ā Ā else {
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(node, adj2, visited2);
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop();
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā scc++;
Ā 
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (scc > 1)
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return false;
Ā Ā Ā Ā Ā Ā Ā Ā }
Ā Ā Ā Ā }
Ā 
Ā Ā Ā Ā return true;
}
Ā 
// Driver code
int main()
{
Ā Ā Ā Ā int N = 2;
Ā Ā Ā Ā vector<vector<int> > edges = { { 0, 1 }, { 1, 0 } };
Ā 
Ā Ā Ā Ā // Function call
Ā Ā Ā Ā bool result = isTourPossible(N, edges);
Ā 
Ā Ā Ā Ā if (result) {
Ā Ā Ā Ā Ā Ā Ā Ā cout << "True" << endl;
Ā Ā Ā Ā }
Ā Ā Ā Ā else {
Ā Ā Ā Ā Ā Ā Ā Ā cout << "False" << endl;
Ā Ā Ā Ā }
Ā Ā Ā Ā return 0;
}
 

















Java


















 






 




 



























// Java program to implement the approach


import java.io.*;


import java.util.*;


Ā 


// Function to find components


class GFG


{


Ā Ā Ā 


Ā Ā Ā Ā // Function to find components


Ā Ā Ā Ā static void


Ā Ā Ā Ā randomOrderDfs(int i,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā ArrayList<ArrayList<Integer> > adj,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā boolean[] visited, Stack<Integer> stk)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj.get(i).size(); j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj.get(i).get(j);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (visited[child] == false) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(child, adj, visited, stk);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā stk.push(i);


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // Function to traverse in the reversed graph


Ā Ā Ā Ā static void dfs2(int i,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā ArrayList<ArrayList<Integer> > adj,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā boolean[] visited)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj.get(i).size(); j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj.get(j).get(0);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (visited[child] == false) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(child, adj, visited);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // Function to check if it is possible


Ā Ā Ā Ā // to reach all other nodes from any node


Ā Ā Ā Ā static boolean isTourPossible(int n, int[][] roads)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā // adj1 stores adjacency matrix of


Ā Ā Ā Ā Ā Ā Ā Ā // original graph adj2 stores


Ā Ā Ā Ā Ā Ā Ā Ā // adjacency matrix of original graph


Ā Ā Ā Ā Ā Ā Ā Ā // by reversing direction of all edges


Ā Ā Ā Ā Ā Ā Ā Ā ArrayList<ArrayList<Integer> > adj1


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = new ArrayList<>();


Ā Ā Ā Ā Ā Ā Ā Ā ArrayList<ArrayList<Integer> > adj2


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = new ArrayList<>();


Ā Ā Ā Ā Ā Ā Ā Ā ;


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj1.add(new ArrayList<Integer>());


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2.add(new ArrayList<Integer>());


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Create graph


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj1.get(roads[i][0]).add(roads[i][1]);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā boolean[] visited1 = new boolean[n];


Ā Ā Ā Ā Ā Ā Ā Ā boolean[] visited2 = new boolean[n];


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā visited1[i] = false;


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā visited2[i] = false;


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā Stack<Integer> stk = new Stack<Integer>();


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Random dfs and maintain stack


Ā Ā Ā Ā Ā Ā Ā Ā // at backtracking (endpoint)


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (visited1[i] == false) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(i, adj1, visited1, stk);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Reverse all the edges


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj1.get(i).size(); j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj1.get(i).get(j);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2.get(child).add(i);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // scc for counting the number of


Ā Ā Ā Ā Ā Ā Ā Ā // strongly connected component


Ā Ā Ā Ā Ā Ā Ā Ā int scc = 0;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Make second dfs at order of stk


Ā Ā Ā Ā Ā Ā Ā Ā while (stk.size() > 0) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int node = stk.peek();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (visited2[node]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(node, adj2, visited2);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā scc++;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (scc > 1)


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return false;


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā return true;


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // Driver code


Ā Ā Ā Ā public static void main(String[] args)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā int N = 2;


Ā Ā Ā Ā Ā Ā Ā Ā int edges[][] = { { 0, 1 }, { 1, 0 } };


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Function call


Ā Ā Ā Ā Ā Ā Ā Ā boolean result = isTourPossible(N, edges);


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā if (result) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("True");


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā else {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā System.out.println("False");


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


}


Ā 


// This code is contributed by garg28harsh.








































Python3


















 






 




 



























# Python code to implement above approach


Ā 


# Function to find components


def randomOrderDfs(i, adj, visited, stk):


Ā Ā Ā Ā visited[i] = True


Ā Ā Ā Ā for child in adj[i]:


Ā Ā Ā Ā Ā Ā Ā Ā if(visited[child] == False):


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(child, adj, visited, stk)


Ā Ā Ā Ā stk.append(i)


Ā 


# Function to traverse in the reversed graph


def dfs2(i, adj, visited):


Ā Ā Ā Ā visited[i] = True


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā for child in adj[i]:


Ā Ā Ā Ā Ā Ā Ā Ā if(visited[child] == False):


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(child, adj, visited)


Ā Ā Ā Ā Ā 


# Function to check if it is possible


# to reach all other nodes from any node


def isTourPossible(n,roads):


Ā Ā Ā Ā # adj1 stores adjacency matrix of


Ā Ā Ā Ā # original graph adj2 stores


Ā Ā Ā Ā # adjacency matrix of original graph


Ā Ā Ā Ā # by reversing direction of all edges


Ā Ā Ā Ā adj1 = [[] for i in range(n)]


Ā Ā Ā Ā adj2 = [[] for i in range(n)]


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā # Create graph


Ā Ā Ā Ā for i in roads:


Ā Ā Ā Ā Ā Ā Ā Ā adj1[i[0]].append(i[1])


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā visited1 = [False]*n


Ā Ā Ā Ā visited2 = [False]*n


Ā Ā Ā Ā stk = []


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā # Random dfs and maintain stack


Ā Ā Ā Ā # at backtracking (endpoint)


Ā Ā Ā Ā for i in range(n):


Ā Ā Ā Ā Ā Ā Ā Ā if(visited1[i]==False):


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(i,adj1,visited1,stk)


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā # Reverse all the edges


Ā Ā Ā Ā for i in range(n):


Ā Ā Ā Ā Ā Ā Ā Ā for child in adj1[i]:


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2[child].append(i)


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā # scc for counting the number of


Ā Ā Ā Ā # strongly connected component


Ā Ā Ā Ā scc=0


Ā Ā Ā Ā Ā 


Ā Ā Ā Ā # Make second dfs at order of stk


Ā Ā Ā Ā while(len(stk) > 0):


Ā Ā Ā Ā Ā Ā Ā Ā node = stk[len(stk)-1]


Ā Ā Ā Ā Ā Ā Ā Ā if(visited2[node]):


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop()


Ā Ā Ā Ā Ā Ā Ā Ā else:


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(node,adj2,visited2)


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.pop()


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā scc = scc + 1


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā 


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if(scc>1):


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return False


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā 


Ā Ā Ā Ā return True


Ā 


# Driver code


N = 2


edges = [[0,1],[1,0]]


Ā 


# Function call


result = isTourPossible(N,edges)


if(result):


Ā Ā Ā Ā print("True")


else:


Ā Ā Ā Ā print("False")


Ā Ā Ā Ā Ā 


# This code is contributed by Pushpesh Raj.








































C#


















 






 




 



























// C# program to implement the approach


using System;


using System.Collections.Generic;


Ā 


// Function to find components


class Program {


Ā Ā // Function to find components


Ā Ā Ā Ā static void RandomOrderDfs(int i, List<List<int> > adj,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā bool[] visited,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Stack<int> stk)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj[i].Count; j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj[i][j];


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (!visited[child]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā RandomOrderDfs(child, adj, visited, stk);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā stk.Push(i);


Ā Ā Ā Ā }


Ā Ā Ā Ā // Function to traverse in the reversed graph


Ā Ā Ā Ā static void Dfs2(int i, List<List<int> > adj,


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā bool[] visited)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj[i].Count; j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj[j][0];


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (!visited[child]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Dfs2(child, adj, visited);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā Ā Ā Ā // Function to check if it is possible


Ā Ā Ā Ā // to reach all other nodes from any node


Ā Ā Ā Ā static bool IsTourPossible(int n, int[][] roads)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā // adj1 stores adjacency matrix of


Ā Ā Ā Ā Ā Ā Ā Ā // original graph adj2 stores


Ā Ā Ā Ā Ā Ā Ā Ā // adjacency matrix of original graph


Ā Ā Ā Ā Ā Ā Ā Ā // by reversing direction of all edges


Ā Ā Ā Ā Ā Ā Ā Ā List<List<int> > adj1 = new List<List<int> >();


Ā Ā Ā Ā Ā Ā Ā Ā List<List<int> > adj2 = new List<List<int> >();


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj1.Add(new List<int>());


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2.Add(new List<int>());


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj1[roads[i][0]].Add(roads[i][1]);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā bool[] visited1 = new bool[n];


Ā Ā Ā Ā Ā Ā Ā Ā bool[] visited2 = new bool[n];


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā visited1[i] = false;


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā visited2[i] = false;


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā Stack<int> stk = new Stack<int>();


Ā Ā Ā Ā Ā Ā Ā 


Ā Ā Ā Ā Ā Ā Ā Ā // Random dfs and maintain stack


Ā Ā Ā Ā Ā Ā Ā Ā // at backtracking (endpoint)


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (!visited1[i]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā RandomOrderDfs(i, adj1, visited1, stk);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā // Reverse all the edges


Ā Ā Ā Ā Ā Ā Ā Ā for (int i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā for (int j = 0; j < adj1[i].Count; j++) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int child = adj1[i][j];


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2[child].Add(i);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā int scc = 0;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā while (stk.Count > 0) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā int node = stk.Peek();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (visited2[node]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.Pop();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā else {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Dfs2(node, adj2, visited2);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā stk.Pop();


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā scc++;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (scc > 1) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return false;


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā return true;


Ā Ā Ā Ā }


Ā Ā Ā Ā // Driver code


Ā Ā Ā Ā static void Main(string[] args)


Ā Ā Ā Ā {


Ā Ā Ā Ā Ā Ā Ā Ā int N = 2;


Ā Ā Ā Ā Ā Ā Ā Ā int[][] edges = new int[][] { new int[] { 0, 1 },


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā new int[] { 1, 0 } };


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā bool result = IsTourPossible(N, edges);


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā Console.WriteLine(result ? "True" : "False");


Ā Ā Ā Ā }


}








































Javascript


















 






 




 



























// JavaScript code to implement above approach


Ā 


function randomOrderDfs(i, adj, visited, stk) {


Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā for (let child of adj[i]) {


Ā Ā Ā Ā Ā Ā Ā Ā if (!visited[child]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(child, adj, visited, stk);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā stk.push(i);


}


Ā 


function dfs2(i, adj, visited) {


Ā Ā Ā Ā visited[i] = true;


Ā 


Ā Ā Ā Ā for (let child of adj[i]) {


Ā Ā Ā Ā Ā Ā Ā Ā if (!visited[child]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(child, adj, visited);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


}


Ā 


function isTourPossible(n, roads) {


Ā Ā Ā Ā // adj1 stores adjacency matrix of


Ā Ā Ā Ā // original graph adj2 stores


Ā Ā Ā Ā // adjacency matrix of original graph


Ā Ā Ā Ā // by reversing direction of all edges


Ā Ā Ā Ā let adj1 = new Array(n), adj2 = new Array(n);


Ā Ā Ā Ā for (let i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā adj1[i] = [];


Ā Ā Ā Ā Ā Ā Ā Ā adj2[i] = [];


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // Create graph


Ā Ā Ā Ā for (let i of roads) {


Ā Ā Ā Ā Ā Ā Ā Ā adj1[i[0]].push(i[1]);


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā let visited1 = new Array(n).fill(false), visited2 = new Array(n).fill(false);


Ā Ā Ā Ā let stk = [];


Ā 


Ā Ā Ā Ā // Random dfs and maintain stack


Ā Ā Ā Ā // at backtracking (endpoint)


Ā Ā Ā Ā for (let i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā if (!visited1[i]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā randomOrderDfs(i, adj1, visited1, stk);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // Reverse all the edges


Ā Ā Ā Ā for (let i = 0; i < n; i++) {


Ā Ā Ā Ā Ā Ā Ā Ā for (let child of adj1[i]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā adj2[child].push(i);


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā // scc for counting the number of


Ā Ā Ā Ā // strongly connected component


Ā Ā Ā Ā let scc = 0;


Ā 


Ā Ā Ā Ā // Make second dfs at order of stk


Ā Ā Ā Ā while (stk.length > 0) {


Ā Ā Ā Ā Ā Ā Ā Ā let node = stk.pop();


Ā Ā Ā Ā Ā Ā Ā Ā if (!visited2[node]) {


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā dfs2(node, adj2, visited2);


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā scc++;


Ā 


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā if (scc > 1)


Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā return false;


Ā Ā Ā Ā Ā Ā Ā Ā }


Ā Ā Ā Ā }


Ā 


Ā Ā Ā Ā return true;


}


Ā 


let N = 2;


let edges = [ [0, 1], [1, 0] ];


Ā 


// Function call


let result = isTourPossible(N, edges);


Ā 


if (result) {


Ā Ā Ā Ā console.log("True");


}


else {


Ā Ā Ā Ā console.log("False");


}


// this code is contributed by dany










































Output


True




Time Complexity: O(N+E), where E is the number of edges
Auxiliary Space: O(N+E)







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