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

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:

1. The implementation below is using BFS.
2. Create a queue and a visited array initially filled with 0, of size V where V is a number of vertices.
3. Insert the starting node in the queue, i.e. push u in the queue and mark u as visited.
4. Run a loop until the queue is not empty.
5. 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.
6. 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#include using namespace std; // This class represents a directed graph using adjacency list // representationclass Graph{    int V;    // No. of vertices    list *adj;    // Pointer to an array containing adjacency listspublic:    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[V];} void Graph::addEdge(int v, int w){    adj[v].push_back(w); // Add w to vâ€™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 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::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 classint 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// representationclass Graph{    private int V;   // No. of vertices    private LinkedList adj[]; //Adjacency List     //Constructor    Graph(int v)    {        V = v;        adj = new LinkedList[v];        for (int i=0; itemp;         // Mark all the vertices as not visited(By default set        // as false)        boolean visited[] = new boolean[V];         // Create a queue for BFS        LinkedList queue = new LinkedList();         // 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 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 collections import defaultdict  #This class represents a directed graph using adjacency list representationclass 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 i in 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 diagramg = 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 = 1if 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// representationclass Graph{  private int V; // No. of vertices  private LinkedList[] 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].AddLast(w);  }   // prints BFS traversal from a given source s  bool isReachable(int s, int d)  {    // LinkedList temp = new LinkedList();     // Mark all the vertices as not visited(By default set    // as false)    bool[] visited = new bool[V];     // Create a queue for BFS    LinkedList queue = new LinkedList();     // 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



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:

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

Implementation:

## C++14

 #include using namespace std;typedef long long ll; vector 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 members = { 2, 5, 7, 9 };     int E = 4;    vector > 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 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 typing import List, Tuple def dfs(start: int, end: int, visited: List[bool], V: int) -> bool:    if start == end:        return True    visited[start] = True    for x in 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: i for i, member in enumerate(members)}     adj = [[] for _ in range(V)]    for a, b in 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[] adj = new LinkedList[100000];         static bool dfs(int start, int end)    {        if (start == end){            return true;        }        visited[start] = true;                 foreach(var x in 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();                 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 (y in 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
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 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 neighbours.
As an exercise, try an extended version of the problem where the complete path between two vertices is also needed.

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