Open In App
Related Articles

Count the number of non-reachable nodes

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Given an undirected graph and a set of vertices, we have to count the number of non-reachable nodes from the given head node using a depth-first search.

Consider below undirected graph with two disconnected components:

not-reachable


In this graph, if we consider 0 as a head node, then the node 0, 1 and 2 are reachable. We mark all the reachable nodes as visited. All those nodes which are not marked as visited i.e, node 3 and 4 are non-reachable nodes. Hence their count is 2.

Example:  

Input :   5
0 1
0 2
1 2
3 4
Output : 2

Here we use BFS:

   First we get the src node from user then call for bfs on it for marking all node connected in that component to mark as visited and then calculate the number of unvisited nodes and return the count of it.

Algorithm for 
Input : 5
0 1
0 2
1 2
3 4
Output : 2
lookes like. for 1
vector<bool>vis(n,0);
push(1).
while loop run till queue is not empty.
1. mark 1 as visited while pop vis[1]=1.
2. then push all node that are unvisited till now from adjacency list of 1.
3. do this till queue it empty.
after break of while,
count the number of unvisited node from visited vector and return count.

Code implementation :

C++

#include <bits/stdc++.h>
#include <iostream>
 
using namespace std;
 
int countNotReachusingBFS(vector<int> adj[], int src, int n)
{
    vector<bool> vis(n, 0);
    queue<int> pq;
    pq.push(src);
    while (!pq.empty()) {
        int start = pq.front();
        pq.pop();
        vis[start] = 1;
        for (auto it : adj[start]) {
            if (vis[it])
                continue;
            else {
                pq.push(it);
            }
        }
    }
    int cnt = 0;
    for (int i = 0; i < n; i++) {
        if (!vis[i])
            ++cnt;
    }
    return cnt;
}
 
int main()
{
    // Create a graph given in the above diagram
    int n = 8;
    // Create a graph in the above diagram
    vector<vector<int> > graph{ { 0, 1 }, { 0, 2 },
                                { 1, 2 }, { 3, 4 },
                                { 4, 5 }, { 6, 7 } };
    vector<int> adj[n + 1];
    for (int i = 0; i < graph.size(); i++) {
        adj[graph[i][0]].push_back(graph[i][1]);
        adj[graph[i][1]].push_back(graph[i][0]);
    }
 
    cout << "the number of node that are not reachable "
            "from 1 are :"
         << endl;
    cout << countNotReachusingBFS(adj, 1, n) << endl;
    // code and approach contributed by Sanket Gode.
    return 0;
}

                    

Java

import java.util.*;
 
public class GFG {
      // Function to return ans
    public static int countNotReachUsingBFS(ArrayList<Integer>[] adj, int src, int n) {
          // visited array to mark nodes visited
        // initially all are unvisited
        boolean[] vis = new boolean[n];
        // Queue for BFS
        Queue<Integer> queue = new LinkedList<>();
        // add source node
        queue.add(src);
       
        // Do BFS
        while (!queue.isEmpty()) {
            // current node
            int node = queue.poll();
            // mark it visited
            vis[node] = true;
            // go to it's adj nodes
            for (int it : adj[node]) {
                if (!vis[it]) {
                    // only visit unvisited  nodes
                    queue.add(it);
                }
            }
        }
 
        int cnt = 0;
        // nodes which are unvisited
        // still they are unreachable nodes
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                cnt++;
            }
        }
        return cnt;
    }
 
    public static void main(String[] args) {
        int n = 8;
 
        ArrayList<Integer>[] adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }
 
        int[][] graph = { {0, 1}, {0, 2}, {1, 2}, {3, 4}, {4, 5}, {6, 7} };
         
        // adj list
        for (int i = 0; i < graph.length; i++) {
            int from = graph[i][0];
            int to = graph[i][1];
            adj[from].add(to);
            adj[to].add(from);
        }
 
        System.out.println("The number of nodes that are not reachable from 1 are:");
        //Function call
        System.out.println(countNotReachUsingBFS(adj, 1, n));
    }
}

                    

Python

from collections import deque
 
def countNotReachusingBFS(adj, src, n):
    vis = [False] * n
    pq = deque()
    pq.append(src)
    while pq:
        start = pq.popleft()
        vis[start] = True
        for it in adj[start]:
            if vis[it]:
                continue
            else:
                pq.append(it)
    cnt = 0
    for i in range(n):
        if not vis[i]:
            cnt += 1
    return cnt
 
if __name__ == '__main__':
    # Create a graph given in the above diagram
    n = 8
    # Create a graph in the above diagram
    graph = [[0, 1], [0, 2],
             [1, 2], [3, 4],
             [4, 5], [6, 7]]
    adj = [[] for _ in range(n)]
    for i in range(len(graph)):
        adj[graph[i][0]].append(graph[i][1])
        adj[graph[i][1]].append(graph[i][0])
 
    print("the number of node that are not reachable from 1 are :")
    print(countNotReachusingBFS(adj, 1, n))

                    

C#

using System;
using System.Collections.Generic;
 
class Program
{
    // Function to count the number of nodes that are not reachable from a source node using BFS
    static int CountNotReachableUsingBFS(List<int>[] adj, int src, int n)
    {
        // Array to keep track of visited nodes
        bool[] vis = new bool[n];
        // Queue for BFS traversal
        Queue<int> queue = new Queue<int>();
        // Enqueue the source node
        queue.Enqueue(src);
 
        while (queue.Count > 0)
        {
            int start = queue.Dequeue();
            vis[start] = true;
 
            // Traverse neighbors of the current node
            foreach (var neighbor in adj[start])
            {
                if (!vis[neighbor])
                {
                    // Enqueue unvisited neighbors
                    queue.Enqueue(neighbor);
                }
            }
        }
 
        // Count nodes that are not reachable
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            if (!vis[i])
                count++;
        }
 
        return count;
    }
 
    static void Main()
    {
        int n = 8;
        // Adjacency list representation of the graph
        List<int>[] adj = new List<int>[n + 1];
 
        for (int i = 0; i <= n; i++)
        {
            adj[i] = new List<int>();
        }
 
        // Edges in the graph
        int[,] graph = {
            { 0, 1 }, { 0, 2 },
            { 1, 2 }, { 3, 4 },
            { 4, 5 }, { 6, 7 }
        };
 
        // Populate the adjacency list
        for (int i = 0; i < graph.GetLength(0); i++)
        {
            adj[graph[i, 0]].Add(graph[i, 1]);
            adj[graph[i, 1]].Add(graph[i, 0]);
        }
 
        // Calculate and display the number of nodes not reachable from node 1
        Console.WriteLine("The number of nodes that are not reachable from 1 are:");
        Console.WriteLine(CountNotReachableUsingBFS(adj, 1, n));
    }
}

                    

Javascript

function countNotReachUsingBFS(adj, src, n) {
    const vis = new Array(n).fill(false);
    const queue = [];
 
    queue.push(src);
 
    while (queue.length > 0) {
        const node = queue.shift();
        vis[node] = true;
         
        for (const it of adj[node]) {
            if (!vis[it]) {
                queue.push(it);
            }
        }
    }
 
    let cnt = 0;
 
    for (let i = 0; i < n; i++) {
        if (!vis[i]) {
            cnt++;
        }
    }
 
    return cnt;
}
 
const n = 8;
const adj = new Array(n).fill().map(() => []);
 
const graph = [
    [0, 1], [0, 2], [1, 2], [3, 4], [4, 5], [6, 7]
];
 
for (let i = 0; i < graph.length; i++) {
    const from = graph[i][0];
    const to = graph[i][1];
    adj[from].push(to);
    adj[to].push(from);
}
 
console.log("The number of nodes that are not reachable from 1 are:");
console.log(countNotReachUsingBFS(adj, 1, n));

                    

Output
the number of node that are not reachable from 1 are :
5



Complexity Analysis :

Time Complexity: O(V+E).
Auxiliary Space: O(V).

In the below implementation, DFS is used. We do DFS from a given source. Since the given graph is undirected, all the vertices that belong to the disconnected component are non-reachable nodes. We use the visited array for this purpose, the array which is used to keep track of non-visited vertices in DFS. In DFS, if we start from the head node it will mark all the nodes connected to the head node as visited. Then after traversing the graph, we count the number of nodes that are not marked as visited from the head node.

Implementation:

C++

// C++ program to count non-reachable nodes
// from a given source using DFS.
#include <iostream>
#include <list>
using namespace std;
 
// Graph class represents a directed graph
// using adjacency list representation
class Graph {
    int V; // No. of vertices
 
    // Pointer to an array containing
    // adjacency lists
    list<int>* adj;
 
    // A recursive function used by DFS
    void DFSUtil(int v, bool visited[]);
 
public:
    Graph(int V); // Constructor
 
    // function to add an edge to graph
    void addEdge(int v, int w);
 
    // DFS traversal of the vertices
    // reachable from v
    int countNotReach(int v);
};
 
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.
    adj[w].push_back(v); // Add v to w's list.
}
 
void Graph::DFSUtil(int v, bool visited[])
{
    // Mark the current node as visited and
    // print it
    visited[v] = true;
 
    // Recur for all the vertices adjacent
    // to this vertex
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}
 
// Returns count of not reachable nodes from
// vertex v.
// It uses recursive DFSUtil()
int Graph::countNotReach(int v)
{
    // Mark all the vertices as not visited
    bool* visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
    // Call the recursive helper function
    // to print DFS traversal
    DFSUtil(v, visited);
 
    // Return count of not visited nodes
    int count = 0;
    for (int i = 0; i < V; i++) {
        if (visited[i] == false)
            count++;
    }
    return count;
}
 
int main()
{
    // Create a graph given in the above diagram
    Graph g(8);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(3, 4);
    g.addEdge(4, 5);
    g.addEdge(6, 7);
 
    cout << g.countNotReach(2);
 
    return 0;
}

                    

Java

// Java program to count non-reachable nodes
// from a given source using DFS.
import java.util.*;
  
// Graph class represents a directed graph
// using adjacency list representation
@SuppressWarnings("unchecked")
class Graph{
      
// No. of vertices   
public int V;
  
// Pointer to an array containing
// adjacency lists
public ArrayList []adj;
  
public Graph(int V)
{
    this.V = V;
    adj = new ArrayList[V];
    for(int i = 0; i < V; i++)
    {
        adj[i] = new ArrayList();
    }
}
   
void addEdge(int v, int w)
{
      
    // add w to v’s list.
    adj[v].add(w);
      
    // add v to w's list.
    adj[w].add(v);
}
   
void DFSUtil(int v, boolean []visited)
{
      
    // Mark the current node as visited and
    // print it
    visited[v] = true;
      
    // Recur for all the vertices adjacent
    // to this vertex
    for(int i : (ArrayList<Integer>)adj[v])
    {
        if (!visited[i])
            DFSUtil(i, visited);
    }
}
   
// Returns count of not reachable nodes from
// vertex v.
// It uses recursive DFSUtil()
int countNotReach(int v)
{
      
    // Mark all the vertices as not visited
    boolean []visited = new boolean[V];
      
    for(int i = 0; i < V; i++)
        visited[i] = false;
   
    // Call the recursive helper function
    // to print DFS traversal
    DFSUtil(v, visited);
   
    // Return count of not visited nodes
    int count = 0;
    for(int i = 0; i < V; i++)
    {
        if (visited[i] == false)
            count++;
    }
    return count;
}
  
// Driver Code
public static void main(String []args)
{
      
    // Create a graph given in the above diagram
    Graph g = new Graph(8);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(3, 4);
    g.addEdge(4, 5);
    g.addEdge(6, 7);
   
    System.out.print(g.countNotReach(2));
}
}
 
// This code is contributed by Pratham76

                    

Python3

# Python3 program to count non-reachable
# nodes from a given source using DFS.
 
# Graph class represents a directed graph
# using adjacency list representation
class Graph:
    def __init__(self, V):
        self.V = V
        self.adj = [[] for i in range(V)]
 
    def addEdge(self, v, w):
        self.adj[v].append(w) # Add w to v’s list.
        self.adj[w].append(v) # Add v to w's list.
     
    def DFSUtil(self, v, visited):
         
        # Mark the current node as
        # visited and print it
        visited[v] = True
     
        # Recur for all the vertices
        # adjacent to this vertex
        i = self.adj[v][0]
        for i in self.adj[v]:
            if (not visited[i]):
                self.DFSUtil(i, visited)
     
    # Returns count of not reachable
    # nodes from vertex v.
    # It uses recursive DFSUtil()
    def countNotReach(self, v):
         
        # Mark all the vertices as not visited
        visited = [False] * self.V
     
        # Call the recursive helper
        # function to print DFS traversal
        self.DFSUtil(v, visited)
     
        # Return count of not visited nodes
        count = 0
        for i in range(self.V):
            if (visited[i] == False):
                count += 1
        return count
 
# Driver Code
if __name__ == '__main__':
 
    # Create a graph given in the
    # above diagram
    g = Graph(8)
    g.addEdge(0, 1)
    g.addEdge(0, 2)
    g.addEdge(1, 2)
    g.addEdge(3, 4)
    g.addEdge(4, 5)
    g.addEdge(6, 7)
 
    print(g.countNotReach(2))
 
# This code is contributed by PranchalK

                    

C#

// C# program to count non-reachable nodes
// from a given source using DFS.
using System;
using System.Collections;
using System.Collections.Generic;
 
// Graph class represents a directed graph
// using adjacency list representation
class Graph{
     
// No. of vertices   
public int V;
 
// Pointer to an array containing
// adjacency lists
public ArrayList []adj;
 
public Graph(int V)
{
    this.V = V;
    adj = new ArrayList[V];
    for(int i = 0; i < V; i++)
    {
        adj[i] = new ArrayList();
    }
}
  
void addEdge(int v, int w)
{
     
    // Add w to v’s list.
    adj[v].Add(w);
     
    // Add v to w's list.
    adj[w].Add(v);
}
  
void DFSUtil(int v, bool []visited)
{
     
    // Mark the current node as visited and
    // print it
    visited[v] = true;
     
    // Recur for all the vertices adjacent
    // to this vertex
    foreach(int i in (ArrayList)adj[v])
    {
        if (!visited[i])
            DFSUtil(i, visited);
    }
}
  
// Returns count of not reachable nodes from
// vertex v.
// It uses recursive DFSUtil()
int countNotReach(int v)
{
     
    // Mark all the vertices as not visited
    bool []visited = new bool[V];
     
    for(int i = 0; i < V; i++)
        visited[i] = false;
  
    // Call the recursive helper function
    // to print DFS traversal
    DFSUtil(v, visited);
  
    // Return count of not visited nodes
    int count = 0;
    for(int i = 0; i < V; i++)
    {
        if (visited[i] == false)
            count++;
    }
    return count;
}
 
// Driver Code
static void Main(string []args)
{
     
    // Create a graph given in the above diagram
    Graph g = new Graph(8);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(3, 4);
    g.addEdge(4, 5);
    g.addEdge(6, 7);
  
    Console.Write(g.countNotReach(2));
}
}
 
// This code is contributed by rutvik_56

                    

Javascript

<script>
    // Javascript program to count non-reachable nodes
    // from a given source using DFS.
     
    // Graph class represents a directed graph
    // using adjacency list representation
    let V = 8;
    let adj = [];
    for(let i = 0; i < V; i++)
    {
      adj.push([]);
    }
     
    function addEdge(v, w)
    {
 
        // Add w to v’s list.
        adj[v].push(w);
 
        // Add v to w's list.
        adj[w].push(v);
    }
 
    function DFSUtil(v, visited)
    {
 
        // Mark the current node as visited and
        // print it
        visited[v] = true;
 
        // Recur for all the vertices adjacent
        // to this vertex
        for(let i = 0; i < adj[v].length; i++)
        {
            if (!visited[adj[v][i]])
                DFSUtil(adj[v][i], visited);
        }
    }
 
    // Returns count of not reachable nodes from
    // vertex v.
    // It uses recursive DFSUtil()
    function countNotReach(v)
    {
 
        // Mark all the vertices as not visited
        let visited = new Array(V);
 
        for(let i = 0; i < V; i++)
            visited[i] = false;
 
        // Call the recursive helper function
        // to print DFS traversal
        DFSUtil(v, visited);
 
        // Return count of not visited nodes
        let count = 0;
        for(let i = 0; i < V; i++)
        {
            if (visited[i] == false)
                count++;
        }
        return count;
    }
     
    // Create a graph given in the above diagram
    addEdge(0, 1);
    addEdge(0, 2);
    addEdge(1, 2);
    addEdge(3, 4);
    addEdge(4, 5);
    addEdge(6, 7);
   
    document.write(countNotReach(2));
     
    // This code is contributed by suresh07.
</script>

                    

Output
5



Time Complexity: O(V+E)
Auxiliary Space: O(V)



Last Updated : 30 Nov, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads