Finding the path from one vertex to rest using BFS

0 0 11 minutes read

Given an adjacency list representation of a directed graph, the task is to find the path from source to every other node in the graph using BFS.

Examples:

Input:

Output:
0 <- 2
1 <- 0 <- 2
2
3 <- 1 <- 0 <- 2
4 <- 5 <- 2
5 <- 2
6 <- 2

Approach: In the images shown below:

que[] array stores the vertices reached and we will enqueue a vertex only if it has not been visited and dequeue it once all its child node have been considered.
In order to distinguish whether the node has been visited or not we will put 1 in visited[] array at the respective index to signify it has been visited and if at given index 0 is present it will signify that it has not been visited.
Parent array is to store the parent node of each vertex. For ex. In case of 0 connected to 2, 2 will be the parent node of 0 and we will put 2 at the index 0 in the parent array.

Below is the implementation of the above approach:

C++14

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the path from
// source (s) to destination (d)
void print(vector<int> parent, int s, int d)
{
   
  // The while loop will stop only when the
  // destination and the source node become equal
  while (s != d)
  {
 
      // Print the destination and store the parent
      // of the node in the destination since parent
      // stores the node through which
      // the current node has been reached
      cout << d << " <- ";
      d = parent[d];
  }
  cout << d << endl;
}
 
// Finding Path using BFS ALgorithm
void bfs(vector<vector<int> > adjList, int source, int n)
{
    vector<int> parent(n, 0);
    vector<int> que(n, 0);
 
    int front = -1, rear = -1;
    vector<int> visited(n, 0);
   
    //Arrays.fill(visited, 0);
    visited = 1;
    parent = source;
 
    // To add any non visited node we will increment the rear
    // and add that vertex to the end of the array (enqueuing)
    que[++rear] = source;
    int k;
 
    // The loop will continue till the rear and front are equal
    while (front != rear)
    {
 
        // Here Dequeuing is nothing but to increment the front int
        k = que[++front];
       
        //L<Integer> list = adjList.get(k);
        for (int j:adjList[k])
        {
            if (visited[j] == 0) 
            {
                que[++rear] = j;
                visited[j] = 1;
                parent[j] = k;
            }
        }
    }
 
    // Print the path from source to every other node
    for (k = 0; k < n; k++)
        print(parent, source, k);
}
 
// Driver code
int main()
{
 
    // Adjacency list representation of the graph
    vector<vector<int> > adjList;
 
    // Vertices 1 and 2 have an incoming edge
    // from vertex 0
    adjList.push_back({1, 2});
 
    // Vertex 3 has an incoming edge
    // from vertex 1
    adjList.push_back({3});
 
    // Vertices 0, 5 and 6 have an incoming
    // edge from vertex 2
    adjList.push_back({0, 5, 6});
 
    // Vertices 1 and 4 have an incoming edge
    // from vertex 3
    adjList.push_back({1, 4});
 
    // Vertices 2 and 3 have an incoming edge
    // from vertex 4
    adjList.push_back({2, 3});
 
    // Vertices 4 and 6 have an incoming edge
    // from vertex 5
    adjList.push_back({4, 6});
 
    // Vertex 5 has an incoming edge
    // from vertex 6
    adjList.push_back({5});
    int n = adjList.size();
    int source = 2;
    bfs(adjList, source, n);
}
 
// This code is contributed by mohit kumar 29.

Java

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
 
class GFG 
{
 
    // Function to print the path from
    // source (s) to destination (d)
    static void print(int parent[], int s, int d)
    {
        // The while loop will stop only when the
        // destination and the source node become equal
        while (s != d) {
 
            // Print the destination and store the parent
            // of the node in the destination since parent
            // stores the node through which
            // the current node has been reached
            System.out.print(d + " <- ");
            d = parent[d];
        }
 
        System.out.println(d);
    }
 
    // Finding Path using BFS ALgorithm
    static void bfs(List<List<Integer> > adjList, int source, int n)
    {
        int parent[] = new int[n];
        int que[] = new int[n];
        Arrays.fill(parent, 0);
        Arrays.fill(que, 0);
 
        int front = -1, rear = -1;
        int visited[] = new int[n];
        Arrays.fill(visited, 0);
        visited = 1;
        parent = source;
 
        // To add any non visited node we will increment the rear
        // and add that vertex to the end of the array (enqueuing)
        que[++rear] = source;
 
        int k;
 
        // The loop will continue till the rear and front are equal
        while (front != rear) {
 
            // Here Dequeuing is nothing but to increment the front int
            k = que[++front];
            List<Integer> list = adjList.get(k);
            for (int i = 0; i < list.size(); i++) {
                int j = list.get(i);
                if (visited[j] == 0) {
                    que[++rear] = j;
                    visited[j] = 1;
                    parent[j] = k;
                }
            }
        }
 
        // Print the path from source to every other node
        for (k = 0; k < n; k++)
            print(parent, source, k);
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        // Adjacency list representation of the graph
        List<List<Integer> > adjList = new ArrayList<>();
 
        // Vertices 1 and 2 have an incoming edge
        // from vertex 0
        List<Integer> tmp = new ArrayList<Integer>(Arrays.asList(1, 2));
        adjList.add(tmp);
 
        // Vertex 3 has an incoming edge from vertex 1
        tmp = new ArrayList<Integer>(Arrays.asList(3));
        adjList.add(tmp);
 
        // Vertices 0, 5 and 6 have an incoming
        // edge from vertex 2
        tmp = new ArrayList<Integer>(Arrays.asList(0, 5, 6));
        adjList.add(tmp);
 
        // Vertices 1 and 4 have an incoming edge
        // from vertex 3
        tmp = new ArrayList<Integer>(Arrays.asList(1, 4));
        adjList.add(tmp);
 
        // Vertices 2 and 3 have an incoming edge
        // from vertex 4
        tmp = new ArrayList<Integer>(Arrays.asList(2, 3));
        adjList.add(tmp);
 
        // Vertices 4 and 6 have an incoming edge
        // from vertex 5
        tmp = new ArrayList<Integer>(Arrays.asList(4, 6));
        adjList.add(tmp);
 
        // Vertex 5 has an incoming edge from vertex 6
        tmp = new ArrayList<Integer>(Arrays.asList(5));
        adjList.add(tmp);
 
        int n = adjList.size();
 
        int source = 2;
        bfs(adjList, source, n);
    }
}

Python3

# Python3 implementation of the approach 
 
# Function to print the path from 
# src (s) to destination (d) 
def printfunc(parent, s, d): 
     
    # The while loop will stop only when 
    # the destination and the src node 
    # become equal 
    while s != d: 
 
        # Print the destination and store 
        # the parent of the node in the 
        # destination since parent stores
        # the node through which the current
        # node has been reached 
        print(str(d) + " <-", end = " ") 
        d = parent[d] 
         
    print(d) 
 
# Finding Path using BFS ALgorithm 
def bfs(adjList, src, n): 
     
    parent = [0] * (n) 
    que = [0] * (n) 
     
    front, rear = -1, -1
    visited = [0] * (n) 
    visited[src] = 1
    parent[src] = src
 
    # To add any non visited node we will 
    # increment the rear and add that vertex 
    # to the end of the array (enqueuing) 
    rear += 1
    que[rear] = src 
 
    # The loop will continue till the rear 
    # and front are equal 
    while front != rear: 
 
        # Here Dequeuing is nothing but to
        # increment the front int 
        front += 1
        k = que[front] 
        List = adjList[k] 
        for i in range(0, len(List)): 
            j = List[i]
             
            if visited[j] == 0:
                rear += 1
                que[rear] = j 
                visited[j] = 1
                parent[j] = k 
                 
    # Print the path from src to every 
    # other node 
    for k in range(0, n): 
        printfunc(parent, src, k) 
     
# Driver code 
if __name__ == "__main__": 
 
    # Adjacency list representation
    # of the graph 
    adjList = [] 
 
    # Vertices 1 and 2 have an incoming edge 
    # from vertex 0 
    adjList.append([1, 2]) 
 
    # Vertex 3 has an incoming edge 
    # from vertex 1 
    adjList.append([3]) 
 
    # Vertices 0, 5 and 6 have an incoming 
    # edge from vertex 2 
    adjList.append([0, 5, 6]) 
 
    # Vertices 1 and 4 have an incoming edge 
    # from vertex 3 
    adjList.append([1, 4]) 
 
    # Vertices 2 and 3 have an incoming edge 
    # from vertex 4 
    adjList.append([2, 3]) 
 
    # Vertices 4 and 6 have an incoming edge 
    # from vertex 5 
    adjList.append([4, 6]) 
 
    # Vertex 5 has an incoming edge 
    # from vertex 6 
    adjList.append([5]) 
 
    n = len(adjList) 
 
    src = 2
    bfs(adjList, src, n) 
     
# This code is contributed by Rituraj Jain

C#

using System;
using System.Collections.Generic;
 
class Program
{
  // Function to print the path from
  // source (s) to destination (d)
  static void Print(int[] parent, int s, int d)
  {
     
    // The while loop will stop only when the
    // destination and the source node become equal
    while (s != d)
    {
       
      // Print the destination and store the parent
      // of the node in the destination since parent
      // stores the node through which
      // the current node has been reached
      Console.Write(d + " <- ");
      d = parent[d];
    }
 
    Console.WriteLine(d);
  }
 
  // Finding Path using BFS ALgorithm
  static void Bfs(List<List<int>> adjList, int source, int n)
  {
    int[] parent = new int[n];
    int[] que = new int[n];
    Array.Fill(parent, 0);
    Array.Fill(que, 0);
 
    int front = -1, rear = -1;
    int[] visited = new int[n];
    Array.Fill(visited, 0);
    visited = 1;
    parent = source;
 
    // To add any non visited node we will increment the rear
    // and add that vertex to the end of the array (enqueuing)
    que[++rear] = source;
 
    int k;
 
    // The loop will continue till the rear and front are equal
    while (front != rear)
    {
      // Here Dequeuing is nothing but to increment the front int
      k = que[++front];
      List<int> list = adjList[k];
      for (int i = 0; i < list.Count; i++)
      {
        int j = list[i];
        if (visited[j] == 0)
        {
          que[++rear] = j;
          visited[j] = 1;
          parent[j] = k;
        }
      }
    }
 
    // Print the path from source to every other node
    for (k = 0; k < n; k++)
      Print(parent, source, k);
  }
 
  // Driver code
  static void Main(string[] args)
  {
    // Adjacency list representation of the graph
    List<List<int>> adjList = new List<List<int>>();
 
    // Vertices 1 and 2 have an incoming edge
    // from vertex 0
    List<int> tmp = new List<int>(new int[] { 1, 2 });
    adjList.Add(tmp);
 
    // Vertex 3 has an incoming edge from vertex 1
    tmp = new List<int>(new int[] { 3 });
    adjList.Add(tmp);
 
    // Vertices 0, 5 and 6 have an incoming
    // edge from vertex 2
    tmp = new List<int>(new int[] { 0, 5, 6 });
    adjList.Add(tmp);
 
    // Vertices 1 and 4 have an incoming edge
    // from vertex 3
    tmp = new List<int>(new int[] { 1, 4 });
    adjList.Add(tmp);
    // Vertices 2 and 3 have an incoming edge
    // from vertex 4
    tmp = new  List<int>(new int[] { 2,3});
    adjList.Add(tmp);
 
    // Vertices 4 and 6 have an incoming edge
    // from vertex 5
    tmp = new  List<int>(new int[] { 4,6 });
    adjList.Add(tmp);
 
    // Vertex 5 has an incoming edge from vertex 6
    tmp = new  List<int>(new int[] { 5 });
    adjList.Add(tmp);
 
    int n = adjList.Count;
 
    int source = 2;
    Bfs(adjList, source, n);
  }
}
using System;
using System.Collections.Generic;
 
 
class Program
{
  // Function to print the path from
  // source (s) to destination (d)
  static void Print(int[] parent, int s, int d)
  {
    // The while loop will stop only when the
    // destination and the source node become equal
    while (s != d)
    {
      // Print the destination and store the parent
      // of the node in the destination since parent
      // stores the node through which
      // the current node has been reached
      Console.Write(d + " <- ");
      d = parent[d];
    }
 
    Console.WriteLine(d);
  }
 
  // Finding Path using BFS ALgorithm
  static void Bfs(List<List<int>> adjList, int source, int n)
  {
    int[] parent = new int[n];
    int[] que = new int[n];
    Array.Fill(parent, 0);
    Array.Fill(que, 0);
 
    int front = -1, rear = -1;
    int[] visited = new int[n];
    Array.Fill(visited, 0);
    visited = 1;
    parent = source;
 
    // To add any non visited node we will increment the rear
    // and add that vertex to the end of the array (enqueuing)
    que[++rear] = source;
 
    int k;
 
    // The loop will continue till the rear and front are equal
    while (front != rear)
    {
      // Here Dequeuing is nothing but to increment the front int
      k = que[++front];
      List<int> list = adjList[k];
      for (int i = 0; i < list.Count; i++)
      {
        int j = list[i];
        if (visited[j] == 0)
        {
          que[++rear] = j;
          visited[j] = 1;
          parent[j] = k;
        }
      }
    }
 
    // Print the path from source to every other node
    for (k = 0; k < n; k++)
      Print(parent, source, k);
  }
 
  // Driver code
  static void Main(string[] args)
  {
    // Adjacency list representation of the graph
    List<List<int>> adjList = new List<List<int>>();
 
    // Vertices 1 and 2 have an incoming edge
    // from vertex 0
    List<int> tmp = new List<int>(new int[] { 1, 2 });
    adjList.Add(tmp);
 
    // Vertex 3 has an incoming edge from vertex 1
    tmp = new List<int>(new int[] { 3 });
    adjList.Add(tmp);
 
    // Vertices 0, 5 and 6 have an incoming
    // edge from vertex 2
    tmp = new List<int>(new int[] { 0, 5, 6 });
    adjList.Add(tmp);
 
    // Vertices 1 and 4 have an incoming edge
    // from vertex 3
    tmp = new List<int>(new int[] { 1, 4 });
    adjList.Add(tmp);
    // Vertices 2 and 3 have an incoming edge
    // from vertex 4
    tmp = new  List<int>(new int[] { 2,3});
    adjList.Add(tmp);
 
    // Vertices 4 and 6 have an incoming edge
    // from vertex 5
    tmp = new  List<int>(new int[] { 4,6 });
    adjList.Add(tmp);
 
    // Vertex 5 has an incoming edge from vertex 6
    tmp = new  List<int>(new int[] { 5 });
    adjList.Add(tmp);
 
    int n = adjList.Count;
 
    int source = 2;
    Bfs(adjList, source, n);
  }
}

Javascript

<script>
 
// JavaScript implementation of the approach
     
    // Function to print the path from
    // source (s) to destination (d)
    function print(parent,s,d)
    {
        // The while loop will stop only when the
        // destination and the source node become equal
        while (s != d) {
  
            // Print the destination and store the parent
            // of the node in the destination since parent
            // stores the node through which
            // the current node has been reached
            document.write(d + " <- ");
            d = parent[d];
        }
  
        document.write(d+"<br>");
    }
     
    // Finding Path using BFS ALgorithm
    function bfs( adjList,source,n)
    {
        let parent = new Array(n);
        let que = new Array(n);
        for(let i=0;i<n;i++)
        {
            parent[i]=0;
            que[i]=0;
        }
  
        let front = -1, rear = -1;
        let visited = new Array(n);
        for(let i=0;i<n;i++)
        {
            visited[i]=0;
        }
        visited = 1;
        parent = source;
  
        // To add any non visited node we will increment the rear
        // and add that vertex to the end of the array (enqueuing)
        que[++rear] = source;
  
        let k;
  
        // The loop will continue till the rear 
        // and front are equal
        while (front != rear) {
  
            // Here Dequeuing is nothing but 
            // to increment the front int
            k = que[++front];
            let list = adjList[k];
            for (let i = 0; i < list.length; i++) {
                let j = list[i];
                if (visited[j] == 0) {
                    que[++rear] = j;
                    visited[j] = 1;
                    parent[j] = k;
                }
            }
        }
  
        // Print the path from source to every other node
        for (k = 0; k < n; k++)
            print(parent, source, k);
    }
     
     // Driver code
     
    // Adjacency list representation of the graph
        let adjList = [];
  
        // Vertices 1 and 2 have an incoming edge
        // from vertex 0
        adjList.push([1, 2])
         
  
        // Vertex 3 has an incoming edge from vertex 1
        adjList.push([3])
  
        // Vertices 0, 5 and 6 have an incoming
        // edge from vertex 2
        adjList.push([0, 5, 6])
  
        // Vertices 1 and 4 have an incoming edge
        // from vertex 3
        adjList.push([1, 4])
  
        // Vertices 2 and 3 have an incoming edge
        // from vertex 4
        adjList.push([2, 3])
  
        // Vertices 4 and 6 have an incoming edge
        // from vertex 5
        adjList.push([4, 6])
  
        // Vertex 5 has an incoming edge from vertex 6
        adjList.push([5])
  
        let n = adjList.length;
  
        let source = 2;
        bfs(adjList, source, n);
 
 
// This code is contributed by unknown2108
 
</script>

Output

0 <- 2
1 <- 0 <- 2
2
3 <- 1 <- 0 <- 2
4 <- 5 <- 2
5 <- 2
6 <- 2

Complexity Analysis:

Time Complexity: O(V + E) where V and E are the numbers of vertices and edges in the graph respectively.
Auxiliary Space: O(V + E).

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