Given a graph G, check if it represents a Bus Topology.
A Bus Topology is the one shown in the image below:
Examples:
Input:
Output: YES Input:
Output: NO
A graph of V vertices represents a bus topology if it satisfies the following two conditions:
- Each node except the starting end ending ones has degree 2 while the starting and ending have degree 1.
- No of edges = No of Vertices – 1.
The idea is to traverse the graph and check if it satisfies the above two conditions. If yes, then it represents a Bus Topology.
Below is the implementation of the above approach:
C++
// CPP program to check if the given graph // represents a bus topology #include <bits/stdc++.h> using namespace std;
// A utility function to add an edge in an // undirected graph. void addEdge(vector< int > adj[], int u, int v)
{ adj[u].push_back(v);
adj[v].push_back(u);
} // A utility function to print the adjacency list // representation of graph void printGraph(vector< int > adj[], int V)
{ for ( int v = 0; v < V; ++v) {
cout << "\n Adjacency list of vertex "
<< v << "\n head " ;
for ( auto x : adj[v])
cout << "-> " << x;
printf ( "\n" );
}
} /* Function to return true if the graph represented by the adjacency list represents a bus topology
else return false */
bool checkBusTopologyUtil(vector< int > adj[], int V, int E)
{ // Number of edges should be equal
// to (Number of vertices - 1)
if (E != (V - 1))
return false ;
// a single node is termed as a bus topology
if (V == 1)
return true ;
int * vertexDegree = new int [V + 1];
memset (vertexDegree, 0, sizeof vertexDegree);
// calculate the degree of each vertex
for ( int i = 1; i <= V; i++) {
for ( auto v : adj[i]) {
vertexDegree[v]++;
}
}
// countDegree2 - number of vertices with degree 2
// countDegree1 - number of vertices with degree 1
int countDegree2 = 0, countDegree1 = 0;
for ( int i = 1; i <= V; i++) {
if (vertexDegree[i] == 2) {
countDegree2++;
}
else if (vertexDegree[i] == 1) {
countDegree1++;
}
else {
// if any node has degree other
// than 1 or 2, it is
// NOT a bus topology
return false ;
}
}
// if both necessary conditions as discussed,
// satisfy return true
if (countDegree1 == 2 && countDegree2 == (V - 2)) {
return true ;
}
return false ;
} // Function to check if the graph represents a bus topology void checkBusTopology(vector< int > adj[], int V, int E)
{ bool isBus = checkBusTopologyUtil(adj, V, E);
if (isBus) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
} // Driver code int main()
{ // Graph 1
int V = 5, E = 4;
vector< int > adj1[V + 1];
addEdge(adj1, 1, 2);
addEdge(adj1, 1, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
checkBusTopology(adj1, V, E);
// Graph 2
V = 4, E = 4;
vector< int > adj2[V + 1];
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 2);
checkBusTopology(adj2, V, E);
return 0;
} |
Java
// java program to check if the given graph // represents a bus topology import java.io.*;
import java.util.*;
class GFG
{ // A utility function to add an edge in an
// undirected graph.
static void addEdge(ArrayList<ArrayList<Integer>> adj, int u, int v)
{
adj.get(u).add(v);
adj.get(v).add(u);
}
// A utility function to print the adjacency list
// representation of graph
static void printGraph(ArrayList<ArrayList<Integer>> adj, int V)
{
for ( int v = 0 ; v < V; ++v)
{
System.out.print( "\n Adjacency list of vertex " + v + "\n head " );
for ( int x : adj.get(v))
{
System.out.print( "-> " + x);
}
System.out.println();
}
}
/* Function to return true if the graph represented
by the adjacency list represents a bus topology
else return false */
static boolean checkBusTopologyUtil(ArrayList<ArrayList<Integer>> adj, int V, int E)
{
// Number of edges should be equal
// to (Number of vertices - 1)
if (E != (V - 1 ))
{
return false ;
}
// a single node is termed as a bus topology
if (V == 1 )
{
return true ;
}
int [] vertexDegree = new int [V + 1 ];
// calculate the degree of each vertex
for ( int i = 1 ; i <= V; i++)
{
for ( int v : adj.get(i))
{
vertexDegree[v]++;
}
}
// countDegree2 - number of vertices with degree 2
// countDegree1 - number of vertices with degree 1
int countDegree2 = 0 , countDegree1 = 0 ;
for ( int i = 1 ; i <= V; i++)
{
if (vertexDegree[i] == 2 )
{
countDegree2++;
}
else if (vertexDegree[i] == 1 )
{
countDegree1++;
}
else
{
// if any node has degree other
// than 1 or 2, it is
// NOT a bus topology
return false ;
}
}
// if both necessary conditions as discussed,
// satisfy return true
if (countDegree1 == 2 && countDegree2 == (V - 2 ))
{
return true ;
}
return false ;
}
// Function to check if the graph represents a bus topology
static void checkBusTopology(ArrayList<ArrayList<Integer>> adj, int V, int E)
{
boolean isBus = checkBusTopologyUtil(adj, V, E);
if (isBus)
{
System.out.println( "YES" );
}
else
{
System.out.println( "NO" );
}
}
// Driver code
public static void main (String[] args)
{
// Graph 1
int V = 5 , E = 4 ;
ArrayList<ArrayList<Integer>> adj1=
new ArrayList<ArrayList<Integer>>();
for ( int i = 0 ; i < V + 1 ; i++)
{
adj1.add( new ArrayList<Integer>());
}
addEdge(adj1, 1 , 2 );
addEdge(adj1, 1 , 3 );
addEdge(adj1, 3 , 4 );
addEdge(adj1, 4 , 5 );
checkBusTopology(adj1, V, E);
// Graph 2
V = 4 ;
E = 4 ;
ArrayList<ArrayList<Integer>> adj2 =
new ArrayList<ArrayList<Integer>>();
for ( int i = 0 ; i < (V + 1 ); i++)
{
adj2.add( new ArrayList<Integer>());
}
addEdge(adj2, 1 , 2 );
addEdge(adj2, 1 , 3 );
addEdge(adj2, 3 , 4 );
addEdge(adj2, 4 , 2 );
checkBusTopology(adj2, V, E);
}
} // This code is contributed by rag2127 |
Python3
# Python3 program to check if the given graph # represents a bus topology # A utility function to add an edge in an # undirected graph. def addEdge(adj, u, v):
adj[u].append(v)
adj[v].append(u)
# A utility function to print the adjacency list # representation of graph def printGraph(adj, V):
for v in range (V):
print ( "Adjacency list of vertex " ,v, "\n head " )
for x in adj[v]:
print ( "-> " ,x,end = " " )
printf()
# /* Function to return true if the graph represented # by the adjacency list represents a bus topology # else return false */ def checkBusTopologyUtil(adj, V, E):
# Number of edges should be equal
# to (Number of vertices - 1)
if (E ! = (V - 1 )):
return False
# a single node is termed as a bus topology
if (V = = 1 ):
return True
vertexDegree = [ 0 ] * (V + 1 )
# calculate the degree of each vertex
for i in range (V + 1 ):
for v in adj[i]:
vertexDegree[v] + = 1
# countDegree2 - number of vertices with degree 2
# countDegree1 - number of vertices with degree 1
countDegree2,countDegree1 = 0 , 0
for i in range ( 1 , V + 1 ):
if (vertexDegree[i] = = 2 ):
countDegree2 + = 1
elif (vertexDegree[i] = = 1 ):
countDegree1 + = 1
else :
# if any node has degree other
# than 1 or 2, it is
# NOT a bus topology
return False
# if both necessary conditions as discussed,
# satisfy return true
if (countDegree1 = = 2 and countDegree2 = = (V - 2 )):
return True
return False
# Function to check if the graph represents a bus topology def checkBusTopology(adj, V, E):
isBus = checkBusTopologyUtil(adj, V, E)
if (isBus):
print ( "YES" )
else :
print ( "NO" )
# Driver code # Graph 1 V, E = 5 , 4
adj1 = [[] for i in range (V + 1 )]
addEdge(adj1, 1 , 2 )
addEdge(adj1, 1 , 3 )
addEdge(adj1, 3 , 4 )
addEdge(adj1, 4 , 5 )
checkBusTopology(adj1, V, E) # Graph 2 V, E = 4 , 4
adj2 = [[] for i in range (V + 1 )]
addEdge(adj2, 1 , 2 )
addEdge(adj2, 1 , 3 )
addEdge(adj2, 3 , 4 )
addEdge(adj2, 4 , 2 )
checkBusTopology(adj2, V, E) # This code is contributed by mohit kumar 29 |
C#
// C# program to check if the given graph // represents a bus topology using System;
using System.Collections.Generic;
public class GFG{
// A utility function to add an edge in an
// undirected graph.
static void addEdge(List<List< int >> adj, int u, int v)
{
adj[u].Add(v);
adj[v].Add(u);
}
// A utility function to print the adjacency list
// representation of graph
static void printGraph(List<List< int >> adj, int V)
{
for ( int v = 0; v < V; ++v)
{
Console.WriteLine( "\n Adjacency list of vertex " + v + "\n head " );
foreach ( int x in adj[v])
{
Console.Write( "-> " + x);
}
Console.WriteLine();
}
}
/* Function to return true if the graph represented
by the adjacency list represents a bus topology
else return false */
static bool checkBusTopologyUtil(List<List< int >> adj, int V, int E)
{
// Number of edges should be equal
// to (Number of vertices - 1)
if (E != (V - 1))
{
return false ;
}
// a single node is termed as a bus topology
if (V == 1)
{
return true ;
}
int [] vertexDegree = new int [V + 1];
// calculate the degree of each vertex
for ( int i = 1; i <= V; i++)
{
foreach ( int v in adj[i])
{
vertexDegree[v]++;
}
}
// countDegree2 - number of vertices with degree 2
// countDegree1 - number of vertices with degree 1
int countDegree2 = 0, countDegree1 = 0;
for ( int i = 1; i <= V; i++)
{
if (vertexDegree[i] == 2)
{
countDegree2++;
}
else if (vertexDegree[i] == 1)
{
countDegree1++;
}
else
{
// if any node has degree other
// than 1 or 2, it is
// NOT a bus topology
return false ;
}
}
// if both necessary conditions as discussed,
// satisfy return true
if (countDegree1 == 2 && countDegree2 == (V - 2))
{
return true ;
}
return false ;
}
// Function to check if the graph represents a bus topology
static void checkBusTopology(List<List< int >> adj, int V, int E)
{
bool isBus = checkBusTopologyUtil(adj, V, E);
if (isBus)
{
Console.WriteLine( "YES" );
}
else
{
Console.WriteLine( "NO" );
}
}
// Driver code
static public void Main ()
{
// Graph 1
int V = 5, E = 4;
List<List< int >> adj1 = new List<List< int >>();
for ( int i = 0; i < V + 1; i++)
{
adj1.Add( new List< int >());
}
addEdge(adj1, 1, 2);
addEdge(adj1, 1, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
checkBusTopology(adj1, V, E);
// Graph 2
V = 4;
E = 4;
List<List< int >> adj2 = new List<List< int >>();
for ( int i = 0; i < V + 1; i++)
{
adj2.Add( new List< int >());
}
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 2);
checkBusTopology(adj2, V, E);
}
} // This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // JavaScript program to check if the given graph // represents a bus topology // A utility function to add an edge in an
// undirected graph.
function addEdge(adj,u,v)
{ adj[u].push(v);
adj[v].push(u);
} // A utility function to print the adjacency list
// representation of graph
function printGraph(adj,V)
{ for (let v = 0; v < V; ++v)
{
document.write( "\n Adjacency list of vertex " +
v + "\n head " );
for (let x=0;x<adj[v].length;x++)
{
document.write( "-> " + adj[v][x]);
}
document.write( "<br>" );
}
} /* Function to return true if the graph represented by the adjacency list represents a bus topology
else return false */
function checkBusTopologyUtil(adj,V,E)
{ // Number of edges should be equal
// to (Number of vertices - 1)
if (E != (V - 1))
{
return false ;
}
// a single node is termed as a bus topology
if (V == 1)
{
return true ;
}
let vertexDegree = new Array(V + 1);
for (let i=0;i<vertexDegree.length;i++)
{
vertexDegree[i]=0;
}
// calculate the degree of each vertex
for (let i = 1; i <= V; i++)
{
for (let v=0;v<adj[i].length;v++)
{
vertexDegree[adj[i][v]]++;
}
}
// countDegree2 - number of vertices with degree 2
// countDegree1 - number of vertices with degree 1
let countDegree2 = 0, countDegree1 = 0;
for (let i = 1; i <= V; i++)
{
if (vertexDegree[i] == 2)
{
countDegree2++;
}
else if (vertexDegree[i] == 1)
{
countDegree1++;
}
else
{
// if any node has degree other
// than 1 or 2, it is
// NOT a bus topology
return false ;
}
}
// if both necessary conditions as discussed,
// satisfy return true
if (countDegree1 == 2 && countDegree2 == (V - 2))
{
return true ;
}
return false ;
} // Function to check if the graph represents a bus topology function checkBusTopology(adj,V,E)
{ let isBus = checkBusTopologyUtil(adj, V, E);
if (isBus)
{
document.write( "YES<br>" );
}
else
{
document.write( "NO<br>" );
}
} // Driver code // Graph 1 let V = 5, E = 4;
let adj1=[];
for (let i = 0; i < V + 1; i++)
{
adj1.push([]);
}
addEdge(adj1, 1, 2);
addEdge(adj1, 1, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
checkBusTopology(adj1, V, E);
// Graph 2
V = 4;
E = 4;
let adj2 = [];
for (let i = 0; i < (V + 1); i++)
{
adj2.push([]);
}
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 2);
checkBusTopology(adj2, V, E);
// This code is contributed by patel2127 </script> |
Output
YES NO
Complexity Analysis:
- Time Complexity : O(E), where E is the number of Edges in the graph.
- Auxiliary Space: O(1).