# Find if there exists a path between two nodes in a directed graph

Sign up for FREE 1 month of Kindle and read all our books for free.

Get FREE domain for 1st year and build your brand new site

You are given a directed graph and two vertices on it. Your task is to find if there exists a path between the first vertex to the second vertex or not.

## Example

#### Consider the following graph:

#### Input to construct the above graph:

```
Input:
No. of nodes in graph = 4
No. of edges in graph = 5
Edges:
1 2
2 3
2 4
3 1
3 4
No. of queries = 2
1 4
4 2
Output:
Yes
No
```

#### Explanation:

For the first query, the answer is**Yes**because we can construct a path as shown in the diagram below

**( 1->2->4 )**but for the second query the answer is

**No**because there are no outgoing edges from vertex four.

## Approach

To solve this problem, we can use either

**BFS**(Breadth First Search) or

**DFS**(Depth First Search) to find if there exists a path between two vertices.

**Some important points:**

1. For representing nodes we will use **1-indexing** or in other words the nodes will be numbered from **1 to number_of_nodes.**

2. To represent the graph we will use a **2-D vector** in C++ and we will use the adjacency list representation as it gives us **lower time and space complexity**. We will resize the 2-D vector to contain vectors equal to no. of nodes.

```
std::vector <std::vector<int>> graph;
graph.resize(number_of_nodes);
```

## 1. Using BFS

Here we will use Breadth First Search in the solution.**Take the first vertex as the source in BFS and if the second vertex is found in traversal print Yes otherwise No.**

#### Pseudocode

```
isPathBFS(Graph G, node source, node end):
Queue Q
Q.enqueue( source ) //add the source node to the queue first
```

tag source as visited

while (Q is not empty):
v = Q.dequeue( )

for all neighbours w of v in Graph G
if w is end :
return true

if w is not visited :
Q.enqueue( w )
tag w as visited
return false

#### Example

In the**Graph G**in the image below, we find whether there exists a path between

**node 1**and

**node 6**using BFS. To find if there exists such a path, we will use BFS with

**node 1**as our source and check if

**node 6**exists in our traversal.

#### Step 1

#### Step 2

#### Step 3

#### Step 4

#### Step 5

#### As __node 6__ is in our traversal ( __BFS__ ), therefore we can draw a path from __node 1__ to __node 6.__ ( 1->2->4->6 )

## Solution using BFS

- C++

### C++

```
// C++ program to check if there exists a path between two vertices of a graph.
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
```

//function definition
bool isPathBFS(vector <vector<int>>& graph, int nodes, int source, int end){
//BFS implementation

bool visited[nodes];
for(int i = 0;i<nodes;++i){
visited[i] = false;
}

queue <int> traversal;
traversal.push(source);
visited[source-1] = true;

while(!traversal.empty()){
//dequeue the front element
source = traversal.front();
traversal.pop();

for(int i = 0;i < (int)graph[source-1].size();++i){
if(graph[source-1][i]==end)
return true;

if(visited[graph[source-1][i]-1]==false){

traversal.push(graph[source-1][i]);
visited[graph[source-1][i]-1] = true;
}
}
}

return false;
}

int main(){
//input
int nodes;
cout << "No. of nodes in graph = ";
cin >> nodes;

int edges;
cout << "No. of edges in graph = ";
cin >> edges;

vector <vector<int>> graph;
graph.resize(nodes);

cout << "Edges:\n";
int source,end;

while(edges--){
cin >> source>>end;

//As we are using 1-indexing for nodes
if(source > nodes || end > nodes){
cout << "Invalid nodes.";
return 1;
}

graph[source-1].push_back(end);
}

int query;
cout << "No. of queries = ";
cin >> query;

while(query--){
cin >> source>>end;

//output
if(source > nodes || end > nodes){
cout << "Invalid nodes.\n";
continue;
}

if(isPathBFS(graph,nodes,source,end)){
cout << "Yes"<<"\n";
}else cout << "No"<<"\n";
}
return 0;
}

### Time Complexity *( for adjancency list representation )*

- Worst case time complexity:
`O(V+E)`

- Best case time complexity:
`Î©(1)`

- Space complexity:
`O(V+E)`

## 2. Using DFS

Here we will use Depth First Search in the solution.**Take the first vertex as the source in DFS and if the second vertex is found in traversal print Yes otherwise No.**

#### Pseudocode

```
isPathDFS(Graph G, node source, node end):
Stack S
S.push( source )
```

tag source as visited

while(S is not empty):
v = S.top()
S.pop()

for all neighbours of w of v in Graph G:
if w is end :
return true

if w is not visited :
S.push( w )
tag w as visited

return false

#### Example

In the**Graph G**in the image below, we find whether there exists a path between

**node 1**and

**node 6**using DFS. To find if there exists such a path, we will use DFS with

**node 1**as our source and check if

**node 6**exists in our traversal.

#### Step 1

#### Step 2

#### Step 3

#### Step 4

#### As __node 6__ is in our traversal ( __DFS__ ), therefore we can draw a path from __node 1__ to __node 6.__ ( 1->2->4->6 )

## Solution using DFS

- C++

### C++

```
// C++ program to check if there exists a path between two vertices of a graph.
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
```

//function definition
bool isPathDFS(vector <vector<int>>& graph, int nodes, int source, int end){
//DFS implementation

bool visited[nodes];
for(int i = 0;i<nodes;++i){
visited[i] = false;
}

stack <int> traversal;
traversal.push(source);
visited[source-1] = true;

while(!traversal.empty()){
//Pop the front element
source = traversal.top();
traversal.pop();

for(int i = 0;i < (int)graph[source-1].size();++i){
if(graph[source-1][i]==end)
return true;

if(visited[graph[source-1][i]-1]==false){

traversal.push(graph[source-1][i]);
visited[graph[source-1][i]-1] = true;
}
}
}

return false;
}

int main(){
//input
int nodes;
cout << "No. of nodes in graph = ";
cin >> nodes;

int edges;
cout << "No. of edges in graph = ";
cin >> edges;

vector <vector<int>> graph;
graph.resize(nodes);

cout << "Edges:\n";
int source,end;

while(edges--){
cin >> source>>end;

//As we are using 1-indexing for nodes
if(source > nodes || end > nodes){
cout << "Invalid nodes.";
return 1;
}

graph[source-1].push_back(end);
}

int query;
cout << "No. of queries = ";
cin >> query;

while(query--){
cin >> source>>end;

//output
if(source > nodes || end > nodes){
cout << "Invalid nodes.\n";
continue;
}

if(isPathDFS(graph,nodes,source,end)){
cout << "Yes"<<"\n";
}else cout << "No"<<"\n";
}
return 0;
}

### Time Complexity *( for adjancency list representation )*

- Worst case time complexity:
`O(V+E)`

- Best case time complexity:
`Î©(1)`

- Space complexity:
`O(V+E)`

#### Tradeoffs between BFS and DFS approach

BFS can be useful to find the**minimum number of edges between two nodes**while DFS may not always give us the path with minimum number of edges as it may traverse one adjacent node very deeply before going to other neighbouring nodes (

*as BFS works level by level while DFS works depth wise*).

For other differences between BFS vs DFS refer to the article :

**BFS vs DFS**

**Exercise: Modify the above algorithm to find any path between two vertices if it exists.**

**Also feel free to share your approach in the discussion thread below. Happy Coding :)**