Check if the given graph represents a Star Topology

Given a graph G, the task is to check if it represents a Star Topology.

A Star Topology is the one shown in the image below:

Examples:



Input : Graph = 
 
Output : YES

Input : Graph =

Output : NO

A graph of V vertices represents a star topology if it satisfies the following three conditions:

  1. One node (also called the central node) has degree V – 1.
  2. All nodes except the central node have degree 1.
  3. No of edges = No of Vertices – 1.

The idea is to traverse the graph and check if it satisfies the above three conditions. If yes, then it represents a Star Topology.

Below is the implementation of the above approach:

CPP

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// CPP program to check if the given graph
// represents a Star 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 Star topology 
   else return false */
bool checkStarTopologyUtil(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 star topology
    // having only a central node
    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]++;
        }
    }
  
    // countCentralNodes strores the count of nodes 
    // with degree V - 1, which should be equal to 1
    // in case of star topology
    int countCentralNodes = 0, centralNode = 0;
  
    for (int i = 1; i <= V; i++) {
        if (vertexDegree[i] == (V - 1)) {
            countCentralNodes++;
            // Store the index of the central node
            centralNode = i;
        }
    }
  
    // there should be only one central node
    // in the star topology
    if (countCentralNodes != 1)
        return false;
  
    for (int i = 1; i <= V; i++) {
        // except for the central node 
        // check if all other nodes have
        // degree 1, if not return false
        if (i == centralNode)
            continue;
        if (vertexDegree[i] != 1) {
            return false;
        }
    }
  
    // if all three necessary
    // conditions as discussed,
    // satisfy return true
    return true;
}
  
// Function to check if the graph 
// represents a Star topology
void checkStarTopology(vector<int> adj[], int V, int E)
{
    bool isStar = checkStarTopologyUtil(adj, V, E);
    if (isStar) {
        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, 1, 4);
    addEdge(adj1, 1, 5);
    checkStarTopology(adj1, V, E);
  
    // Graph 2
    V = 5, E = 4;
    vector<int> adj2[V + 1];
    addEdge(adj2, 1, 2);
    addEdge(adj2, 1, 3);
    addEdge(adj2, 3, 4);
    addEdge(adj2, 4, 5);
    checkStarTopology(adj2, V, E);
  
    return 0;
}

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Python3

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# Python3 program to check if the given graph
# represents a star 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 prthe 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 star topology
#    else return false */
def checkStarTopologyUtil(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
  
    # countCentralNodes strores the count of nodes
    # with degree V - 1, which should be equal to 1
    # in case of star topology
    countCentralNodes = 0
    centralNode = 0
  
    for i in range(1, V + 1):
        if (vertexDegree[i] == (V - 1)):
            countCentralNodes += 1
            # Store the index of the central node
            centralNode = i
  
    # there should be only one central node
    # in the star topology
    if (countCentralNodes != 1):
        return False
  
    for i in range(1, V + 1):
        # except for the central node
        # check if all other nodes have
        # degree 1, if not return false
        if (i == centralNode):
            continue
        if (vertexDegree[i] != 1):
            return False
  
    # if all three necessary
    # conditions as discussed,
    # satisfy return true
    return True
  
# Function to check if the graph represents a bus topology
def checkStarTopology(adj, V, E):
  
    isStar = checkStarTopologyUtil(adj, V, E)
    if (isStar):
        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, 1, 4)
addEdge(adj1, 1, 5)
checkStarTopology(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)
checkStarTopology(adj2, V, E)
  
# This code is contribued by mohit kumar 29

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Output:

YES
NO

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




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Improved By : mohit kumar 29