Find if there is a path between two vertices in a directed graph

Last Updated : 07 Feb, 2023

Given a Directed Graph and two vertices in it, check whether there is a path from the first given vertex to second.
Example:

Consider the following Graph:


Input : (u, v) = (1, 3)
Output: Yes
Explanation: There is a path from 1 to 3, 1 -> 2 -> 3

Input : (u, v) = (3, 6)
Output: No
Explanation: There is no path from 3 to 6

Approach: Either Breadth First Search (BFS) or Depth First Search (DFS) can be used to find path between two vertices. Take the first vertex as a source in BFS (or DFS), follow the standard BFS (or DFS). If the second vertex is found in our traversal, then return true else return false.

BFS Algorithm:

The implementation below is using BFS.
Create a queue and a visited array initially filled with 0, of size V where V is a number of vertices.
Insert the starting node in the queue, i.e. push u in the queue and mark u as visited.
Run a loop until the queue is not empty.
Dequeue the front element of the queue. Iterate all its adjacent elements. If any of the adjacent elements is the destination return true. Push all the adjacent and unvisited vertices in the queue and mark them as visited.
Return false as the destination is not reached in BFS.

Implementation: C++, Java, and Python codes that use BFS for finding the reachability of the second vertex from the first vertex.

C++

// C++ program to check if there is exist a path between two vertices
// of a graph.
#include<iostream>
#include <list>
using namespace std;
 
// This class represents a directed graph using adjacency list
// representation
class Graph
{
    int V;   // No. of vertices
    list<int> *adj;   // Pointer to an array containing adjacency lists
public:
    Graph(int V); // Constructor
    void addEdge(int v,int w);// function to add an edge to graph
    bool isReachable(int s,int d); 
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj =new list<int>[V];
}
 
void Graph::addEdge(int v,int w)
{
    adj[v].push_back(w);// Add w to v’s list.
}
 
// A BFS based function to check whether d is reachable from s.
bool Graph::isReachable(int s,int d)
{
    // Base case
    if (s == d)
      return true;
 
    // Mark all the vertices as not visited
    bool *visited =new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] =false;
 
    // Create a queue for BFS
    list<int> queue;
 
    // Mark the current node as visited and enqueue it
    visited[s] =true;
    queue.push_back(s);
 
    // it will be used to get all adjacent vertices of a vertex
    list<int>::iterator i;
 
    while (!queue.empty())
    {
        // Dequeue a vertex from queue and print it
        s = queue.front();
        queue.pop_front();
 
        // Get all adjacent vertices of the dequeued vertex s
        // If a adjacent has not been visited, then mark it visited
        // and enqueue it
        for (i = adj[s].begin(); i != adj[s].end(); ++i)
        {
            // If this adjacent node is the destination node, then
            // return true
            if (*i == d)
                return true;
 
            // Else, continue to do BFS
            if (!visited[*i])
            {
                visited[*i] =true;
                queue.push_back(*i);
            }
        }
    }
     
    // If BFS is complete without visiting d
    return false;
}
 
// Driver program to test methods of graph class
int main()
{
    // Create a graph given in the above diagram
    Graph g(4);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 3);
 
    int u = 1, v = 3;
    if(g.isReachable(u, v))
        cout<<"\n There is a path from " << u <<" to " << v;
    else
        cout<<"\n There is no path from " << u <<" to " << v;
 
    u = 3, v = 1;
    if(g.isReachable(u, v))
        cout<<"\n There is a path from " << u <<" to " << v;
    else
        cout<<"\n There is no path from " << u <<" to " << v;
 
    return 0;
}

Java

// Java program to check if there is exist a path between two vertices
// of a graph.
import java.io.*;
import java.util.*;
import java.util.LinkedList;
 
// This class represents a directed graph using adjacency list
// representation
class Graph
{
    private int V;  // No. of vertices
    private LinkedList<Integer> adj[];//Adjacency List
 
    //Constructor
    Graph(int v)
    {
        V = v;
        adj =new LinkedList[v];
        for (int i=0; i<v; ++i)
            adj[i] =new LinkedList();
    }
 
    //Function to add an edge into the graph
    void addEdge(int v,int w)  {   adj[v].add(w);   }
 
    //prints BFS traversal from a given source s
    Boolean isReachable(int s,int d)
    {
        LinkedList<Integer>temp;
 
        // Mark all the vertices as not visited(By default set
        // as false)
        boolean visited[] =new boolean[V];
 
        // Create a queue for BFS
        LinkedList<Integer> queue =new LinkedList<Integer>();
 
        // Mark the current node as visited and enqueue it
        visited[s]=true;
        queue.add(s);
 
        // 'i' will be used to get all adjacent vertices of a vertex
        Iterator<Integer> i;
        while (queue.size()!=0)
        {
            // Dequeue a vertex from queue and print it
            s = queue.poll();
 
            int n;
            i = adj[s].listIterator();
 
            // Get all adjacent vertices of the dequeued vertex s
            // If a adjacent has not been visited, then mark it
            // visited and enqueue it
            while (i.hasNext())
            {
                n = i.next();
 
                // If this adjacent node is the destination node,
                // then return true
                if (n==d)
                    return true;
 
                // Else, continue to do BFS
                if (!visited[n])
                {
                    visited[n] =true;
                    queue.add(n);
                }
            }
        }
 
        // If BFS is complete without visited d
        return false;
    }
 
    // Driver method
    public static void main(String args[])
    {
        // Create a graph given in the above diagram
        Graph g =new Graph(4);
        g.addEdge(0,1);
        g.addEdge(0,2);
        g.addEdge(1,2);
        g.addEdge(2,0);
        g.addEdge(2,3);
        g.addEdge(3,3);
 
        int u =1;
        int v =3;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;
 
        u =3;
        v =1;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;
    }
}
// This code is contributed by Aakash Hasija

Python3

# program to check if there is exist a path between two vertices
# of a graph
 
from collectionsimport defaultdict
  
#This class represents a directed graph using adjacency list representation
class Graph:
  
    def __init__(self,vertices):
        self.V= vertices#No. of vertices
        self.graph= defaultdict(list)# default dictionary to store graph
  
    # function to add an edge to graph
    def addEdge(self,u,v):
        self.graph[u].append(v)
      
     # Use BFS to check path between s and d
    def isReachable(self, s, d):
        # Mark all the vertices as not visited
        visited=[False]*(self.V)
  
        # Create a queue for BFS
        queue=[]
  
        # Mark the source node as visited and enqueue it
        queue.append(s)
        visited[s]= True
  
        while queue:
 
            #Dequeue a vertex from queue
            n= queue.pop(0)
             
            # If this adjacent node is the destination node,
            # then return true
            if n== d:
                   return True
 
            #  Else, continue to do BFS
            for iin self.graph[n]:
                if visited[i]== False:
                    queue.append(i)
                    visited[i]= True
         # If BFS is complete without visited d
        return False
  
# Create a graph given in the above diagram
g= Graph(4)
g.addEdge(0,1)
g.addEdge(0,2)
g.addEdge(1,2)
g.addEdge(2,0)
g.addEdge(2,3)
g.addEdge(3,3)
 
u=1; v= 3
 
if g.isReachable(u, v):
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))
 
u= 3; v= 1
if g.isReachable(u, v) :
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))
 
#This code is contributed by Neelam Yadav

C#

// C# program to check if there is
// exist a path between two vertices
// of a graph.
using System;
using System.Collections;
using System.Collections.Generic;
 
// This class represents a directed
// graph using adjacency list
// representation
class Graph
{
  private int V;// No. of vertices
  private LinkedList<int>[] adj;//Adjacency List
 
  // Constructor
  Graph(int v)
  {
    V = v;
    adj =new LinkedList<int>[v];
    for (int i = 0; i < v; ++i)
      adj[i] =new LinkedList<int>();
  }
 
  // Function to add an edge into the graph
  void addEdge(int v,int w)
  {
    adj[v].AddLast(w);
  }
 
  // prints BFS traversal from a given source s
  bool isReachable(int s,int d)
  {
    // LinkedList<int> temp = new LinkedList<int>();
 
    // Mark all the vertices as not visited(By default set
    // as false)
    bool[] visited =new bool[V];
 
    // Create a queue for BFS
    LinkedList<int> queue =new LinkedList<int>();
 
    // Mark the current node as visited and enqueue it
    visited[s] =true;
    queue.AddLast(s);
 
    // 'i' will be used to get all adjacent vertices of a vertex
    IEnumerator i;     
    while (queue.Count != 0)
    {
 
      // Dequeue a vertex from queue and print it
      s = queue.First.Value;
      queue.RemoveFirst();
      int n;
      i = adj[s].GetEnumerator();
 
      // Get all adjacent vertices of the dequeued vertex s
      // If a adjacent has not been visited, then mark it
      // visited and enqueue it
      while (i.MoveNext())
      {
        n = (int)i.Current;
 
        // If this adjacent node is the destination node,
        // then return true
        if (n == d)
          return true;
 
        // Else, continue to do BFS
        if (!visited[n])
        {
          visited[n] =true;
          queue.AddLast(n);
        }
      }
    }
 
    // If BFS is complete without visited d
    return false;
  }
 
  // Driver method
  public static void Main(string[] args)
  {
 
    // Create a graph given in the above diagram
    Graph g =new Graph(4);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 3);
    int u = 1;
    int v = 3;
    if (g.isReachable(u, v))
      Console.WriteLine("There is a path from " + u +" to " + v);
    else
      Console.WriteLine("There is no path from " + u +" to " + v);
    u = 3;
    v = 1;
    if (g.isReachable(u, v))
      Console.WriteLine("There is a path from " + u +" to " + v);
    else
      Console.WriteLine("There is no path from " + u +" to " + v);
  }
}
 
// This code is contributed by sanjeev2552

Javascript

<script>
// Javascript program to check if there is exist a path between two vertices
// of a graph.
let  V;
let adj;
 
function Graph( v)
{
        V = v;
        adj =new Array(v);
        for (let i = 0; i < v; ++i)
            adj[i] = [];
}
 
// Function to add an edge into the graph
function addEdge(v,w)
{
    adj[v].push(w);
}
 
// prints BFS traversal from a given source s
function isReachable(s,d)
{
    let temp;
  
        // Mark all the vertices as not visited(By default set
        // as false)
        let visited =new Array(V);
         for(let i = 0; i < V; i++)
            visited[i] =false;
             
        // Create a queue for BFS
        let queue = [];
  
        // Mark the current node as visited and enqueue it
        visited[s] =true;
        queue.push(s);
  
        while (queue.length != 0)
        {
            // Dequeue a vertex from queue and print it
            n = queue.shift();
              
            if(n == d)
                return true;
            for(let i = 0; i < adj[n].length; i++)
            {
                if (visited[adj[n][i]] ==false)
                {
                    queue.push(adj[n][i]);
                    visited[adj[n][i]] =true;
                }
            }
             
        }
  
        // If BFS is complete without visited d
        return false;
}
 
// Driver method
Graph(4);
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 2);
addEdge(2, 0);
addEdge(2, 3);
addEdge(3, 3);
 
let u = 1;
let v = 3;
if (isReachable(u, v))
    document.write("There is a path from " + u +" to " + v+"<br>");
else
    document.write("There is no path from " + u +" to " + v+"<br>");
 
u = 3;
v = 1;
if (isReachable(u, v))
    document.write("There is a path from " + u +" to " + v+"<br>");
else
    document.write("There is no path from " + u +" to " + v+"<br>");
 
// This code is contributed by avanitrachhadiya2155
</script>

Output

 There is a path from 1 to 3
 There is no path from 3 to 1

Complexity Analysis:

Time Complexity: O(V+E) where V is number of vertices in the graph and E is number of edges in the graph.
Space Complexity: O(V).
There can be atmost V elements in the queue. So the space needed is O(V).

DFS Algorithm:

if start==end return 1 since we have to reached our destination.
Mark start as visited.
Traverse directly connected vertices of start and recur the function dfs for every such unexplored vertex.
return 0 if we do not reach our destination.

Implementation:

C++14

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
vector<ll> adj[100000];
bool visited[100000];
 
bool dfs(int start,int end)
{
    if (start == end)
        return true;
    visited[start] = 1;
    for (auto x : adj[start]) {
        if (!visited[x])
            if (dfs(x, end))
                return true;
    }
    return false;
}
 
int main()
{
 
    int V = 4;
    vector<ll> members = { 2, 5, 7, 9 };
 
    int E = 4;
    vector<pair<ll, ll> > connections
        = { { 2, 9 }, { 7, 2 }, { 7, 9 }, { 9, 5 } };
 
    for (int i = 0; i < E; i++)
        adj[connections[i].first].push_back(
            connections[i].second);
 
    int sender = 7, receiver = 9;
 
    if (dfs(sender, receiver))
        cout <<"1";
    else
        cout <<"0";
 
    return 0;
}
// this code is contributed by prophet1999

Java

/*package whatever //do not write package name here */
import java.util.*;
 
public class GFG {
     
    public static Vector<Integer> adj[] =new Vector[100000];
    public static int[] visited =new int[100000];
     
    static boolean dfs(int start,int end)
    {
        if (start == end){
            return true;
        }
 
        visited[start] =1;
        for(int i =0; i < adj[start].size(); i++){
            int x = adj[start].get(i);
            if (visited[x] ==0){
                if (dfs(x, end)){
                    return true;
                }
            }
        }
        return false;
    }
     
    public static void main(String[] args) {
        int V =4;
        int[] members = {2,5,7,9 };
 
        int E =4;
        int[][] connections = { {2,9 }, {7,2 }, {7,9 }, {9,5 } };
 
        for(int i =0;i<100000;i++) {
            adj[i] =new Vector<>();
        }
        for (int i =0; i < E; i++){
            // System.out.println(connections[i][0] + connections[i][1]);
            adj[connections[i][0]].add(connections[i][1]);
        }
 
        int sender =7;
        int receiver =9;
         
        if (dfs(sender, receiver) ==true){
            System.out.println("1");
        }
        else{
            System.out.println("0");
        }
    }
}
 
// The code is contributed by Gautam goel.

Python3

from typingimport List,Tuple
 
def dfs(start:int, end:int, visited:List[bool], V:int)->bool:
    if start== end:
        return True
    visited[start]= True
    for xin adj[start]:
        if not visited[x]:
            if dfs(x, end, visited, V):
                return True
    return False
 
if __name__== '__main__':
    V= 4
    members= [2,5,7,9]
 
    E= 4
    connections= [ (2,9), (7,2), (7,9), (9,5) ]
 
    member_to_index= {member: ifor i, memberin enumerate(members)}
 
    adj= [[]for _in range(V)]
    for a, bin connections:
        a= member_to_index[a]
        b= member_to_index[b]
        adj[a].append(b)
 
    sender= member_to_index[7]
    receiver= member_to_index[9]
 
    visited= [False]* V
    if dfs(sender, receiver, visited, V):
        print("1")
    else:
        print("0")

C#

using System;
using System.Collections.Generic;
 
public class GFG {
     
    public static bool[] visited =new bool[100000];
    public static LinkedList<int>[] adj =new LinkedList<int>[100000];
     
    static bool dfs(int start,int end)
    {
        if (start == end){
            return true;
        }
        visited[start] =true;
         
        foreach(var xin adj[start]){
             
            if (visited[x] ==false)
                if (dfs(x, end) ==true)
                    return true;
        }
        return false;
    }
     
    static void Main() {
        int V = 4;
         
        int[] members = { 2, 5, 7, 9 };
 
        int E = 4;
        int[,] connections = { { 2, 9 }, { 7, 2 }, { 7, 9 }, { 9, 5 } };
 
        for (int i = 0; i < 100000; i++)
            adj[i] =new LinkedList<int>();
         
        for (int i = 0; i < E; i++){
            adj[connections[i,0]].AddLast(connections[i,1]);
        }
             
        int sender = 7;
        int receiver = 9;
 
        if (dfs(sender, receiver) ==true){
            Console.WriteLine("1");
        }
        else{
            Console.WriteLine("0");
        }
         
    }
}
 
// The code is contributed by Gautam goel.

JavaScript

let adj = Array.from(Array(100000), () =>new Array());
   let visited =new Array(100000);
   function dfs(start, end) {
     if (start == end)return true;
     visited[start] = 1;
     for (yin adj[start]) {
       x = adj[start][y];
       if (!visited[x])if (dfs(x, end))return true;
     }
     return false;
   }
 
   let V = 4;
   members = [2, 5, 7, 9];
 
   let E = 4;
   connections = [
     [2, 9],
     [7, 2],
     [7, 9],
     [9, 5],
   ];
 
   for (let i = 0; i < E; i++)
     adj[connections[i][0]].push(connections[i][1]);
 
   let sender = 7,
     receiver = 9;
 
   if (dfs(sender, receiver)) console.log("1");
   else console.log("0");

Output

Complexity Analysis:

Time Complexity: O(V+E) where V is number of vertices in the graph and E is number of edges in the graph.
Space Complexity: O(V).
There can be atmost V elements in the stack. So the space needed is O(V).

Trade-offs between BFS and DFS:

Breadth-First search can be useful to find the shortest path between nodes, and depth-first search may traverse one adjacent node very deeply before ever going into immediate neigurs.
As an exercise, try an extended version of the problem where the complete path between two vertices is also needed.

Hamiltonian Cycle

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Article Tags :

Practice Tags :

Find if there is a path between two vertices in a directed graph

C++

Java

Python3

C#

Javascript

C++14

Java

Python3

C#

JavaScript

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