Count of nodes with maximum connection in an undirected graph

Given an undirected graph with N nodes and E edges, followed by E edge connections. The task is to find the count of nodes with maximum connections

Examples:

Input: N =10, E =13,  
edges[][] = { {1, 4}, {2, 3}, {4, 5}, {3, 9}, {6, 9}, {3, 8}, {10, 4},  
                    {2, 7}, {3, 6}, {2, 8}, {9, 2}, {1, 10}, {9, 10} }
Output: 3
Explanation: The connections for each of the nodes are show below:

• 1 -> 4, 10
• 2 -> 3, 7, 8, 9
• 3 -> 2, 9, 8, 6
• 4 -> 1, 5, 10
• 5 -> 4
• 6 -> 9, 3
• 7 -> 2
• 8 -> 3, 2
• 9 -> 3, 6, 2, 10
• 10 -> 4, 9, 1

Therefore, number of nodes with maximum connections are 3 viz. {2, 3, 9}

Input: N = 8, E = 7,  
edges[][] = { {0, 2}, {1, 5}, {2, 3}, {5, 7}, {2, 4}, {5, 6}, {1, 2} }
Output: 3

Approach: The task can be solved by storing the number of connected nodes for every node inside a vector. And then, find the maximum connected nodes to any of the nodes & get its count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of nodes
// with maximum connections
void get(map<int, vector<int> > graph)
{
    // Stores the number of connections
    // of each node
    vector<int> v;

    // Stores the maximum connections
    int mx = -1;
    for (int i = 0; i < graph.size(); i++) {
        v.push_back(graph[i].size());
        mx = max(mx, (int)graph[i].size());
    }

    // Resultant count
    int cnt = 0;
    for (auto i : v) {
        if (i == mx)
            cnt++;
    }

    cout << cnt << endl;
}

// Drive Code
int main()
{
    map<int, vector<int> > graph;
    int nodes = 10, edges = 13;

    // 1
    graph[1].push_back(4);
    graph[4].push_back(1);

    // 2
    graph[2].push_back(3);
    graph[3].push_back(2);

    // 3
    graph[4].push_back(5);
    graph[5].push_back(4);

    // 4
    graph[3].push_back(9);
    graph[9].push_back(3);

    // 5
    graph[6].push_back(9);
    graph[9].push_back(6);

    // 6
    graph[3].push_back(8);
    graph[8].push_back(3);

    // 7
    graph[10].push_back(4);
    graph[4].push_back(10);

    // 8
    graph[2].push_back(7);
    graph[7].push_back(2);

    // 9
    graph[3].push_back(6);
    graph[6].push_back(3);

    // 10
    graph[2].push_back(8);
    graph[8].push_back(2);

    // 11
    graph[9].push_back(2);
    graph[2].push_back(9);

    // 12
    graph[1].push_back(10);
    graph[10].push_back(1);

    // 13
    graph[9].push_back(10);
    graph[10].push_back(9);

    get(graph);
    return 0;
}

Java

// Java program for the above approach
import java.util.ArrayList;
import java.util.HashMap;

class GFG {

  // Function to count the number of nodes
  // with maximum connections
  static void get(HashMap<Integer, ArrayList<Integer>> graph)
  {

    // Stores the number of connections
    // of each node
    ArrayList<Integer> v = new ArrayList<Integer>();

    // Stores the maximum connections
    int mx = -1;
    for (int i = 0; i < graph.size(); i++) {
      v.add(graph.get(i).size());
      mx = Math.max(mx, (int) graph.get(i).size());
    }

    // Resultant count
    int cnt = 0;
    for (int i : v) {
      if (i == mx)
        cnt++;
    }

    System.out.println(cnt);
  }

  // Drive Code
  public static void main(String args[]) {

    HashMap<Integer, ArrayList<Integer>> graph = new HashMap<Integer, ArrayList<Integer>>();
    int nodes = 10, edges = 13;

    for (int i = 0; i <= nodes; i++) {
      graph.put(i, new ArrayList<>());
    }
    // 1
    graph.get(1).add(4);
    graph.get(4).add(1);

    // 2
    graph.get(2).add(3);
    graph.get(3).add(2);

    // 3
    graph.get(4).add(5);
    graph.get(5).add(4);

    // 4
    graph.get(3).add(9);
    graph.get(9).add(3);

    // 5
    graph.get(6).add(9);
    graph.get(9).add(6);

    // 6
    graph.get(3).add(8);
    graph.get(8).add(3);

    // 7
    graph.get(10).add(4);
    graph.get(4).add(10);

    // 8
    graph.get(2).add(7);
    graph.get(7).add(2);

    // 9
    graph.get(3).add(6);
    graph.get(6).add(3);

    // 10
    graph.get(2).add(8);
    graph.get(8).add(2);

    // 11
    graph.get(9).add(2);
    graph.get(2).add(9);

    // 12
    graph.get(1).add(10);
    graph.get(10).add(1);

    // 13
    graph.get(9).add(10);
    graph.get(10).add(9);

    get(graph);
  }
}

// This code is contributed by saurabh_jaiswal.

Python3

# Python code for the above approach

# Function to count the number of nodes
# with maximum connections
def get(graph):

    # Stores the number of connections
    # of each node
    v = [];

    # Stores the maximum connections
    mx = -1;
    for arr in graph.values():
        v.append(len(arr));
        mx = max(mx, (len(arr)));

    # Resultant count
    cnt = 0;
    for i in v:
        if (i == mx):
            cnt += 1

    print(cnt)

# Drive Code
graph = {}

nodes = 10
edges = 13;
for i in range(1, nodes + 1):
    graph[i] = []

# 1
graph[1].append(4);
graph[4].append(1);

# 2
graph[2].append(3);
graph[3].append(2);

# 3
graph[4].append(5);
graph[5].append(4);

# 4
graph[3].append(9);
graph[9].append(3);

# 5
graph[6].append(9);
graph[9].append(6);

# 6
graph[3].append(8);
graph[8].append(3);

# 7
graph[10].append(4);
graph[4].append(10);

# 8
graph[2].append(7);
graph[7].append(2);

# 9
graph[3].append(6);
graph[6].append(3);

# 10
graph[2].append(8);
graph[8].append(2);

# 11
graph[9].append(2);
graph[2].append(9);

# 12
graph[1].append(10);
graph[10].append(1);

# 13
graph[9].append(10);
graph[10].append(9);

get(graph);

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG {

  // Function to count the number of nodes
  // with maximum connections
  static void get(Dictionary<int, List<int>> graph)
  {

    // Stores the number of connections
    // of each node
    List<int> v = new List<int>();

    // Stores the maximum connections
    int mx = -1;
    for (int i = 0; i < graph.Count; i++) {
      v.Add(graph[i].Count);
      mx = Math.Max(mx, (int) graph[i].Count);
    }

    // Resultant count
    int cnt = 0;
    foreach (int i in v) {
      if (i == mx)
        cnt++;
    }

    Console.WriteLine(cnt);
  }

  // Drive Code
  public static void Main(String []args) {

    Dictionary<int, List<int>> graph = new Dictionary<int, List<int>>();
    int nodes = 10;

    for (int i = 0; i <= nodes; i++) {
      graph.Add(i, new List<int>());
    }
    // 1
    graph[1].Add(4);
    graph[4].Add(1);

    // 2
    graph[2].Add(3);
    graph[3].Add(2);

    // 3
    graph[4].Add(5);
    graph[5].Add(4);

    // 4
    graph[3].Add(9);
    graph[9].Add(3);

    // 5
    graph[6].Add(9);
    graph[9].Add(6);

    // 6
    graph[3].Add(8);
    graph[8].Add(3);

    // 7
    graph[10].Add(4);
    graph[4].Add(10);

    // 8
    graph[2].Add(7);
    graph[7].Add(2);

    // 9
    graph[3].Add(6);
    graph[6].Add(3);

    // 10
    graph[2].Add(8);
    graph[8].Add(2);

    // 11
    graph[9].Add(2);
    graph[2].Add(9);

    // 12
    graph[1].Add(10);
    graph[10].Add(1);

    // 13
    graph[9].Add(10);
    graph[10].Add(9);

    get(graph);
  }
}

// This code is contributed by shikhasingrajput

Javascript

<script>
       // JavaScript code for the above approach

       // Function to count the number of nodes
       // with maximum connections
       function get(graph)
       {

           // Stores the number of connections
           // of each node
           let v = [];

           // Stores the maximum connections
           let mx = -1;
           for (let [key, arr] of graph) {
               v.push(arr.length);
               mx = Math.max(mx, (arr.length));
           }

           // Resultant count
           let cnt = 0;
           for (let i of v) {
               if (i == mx)
                   cnt++;
           }

           document.write(cnt + "<br>")
       }

       // Drive Code

       let graph = new Map();


       let nodes = 10, edges = 13;
       for (let i = 1; i <= nodes; i++) {
           graph.set(i, []);
       }
       // 1
       graph.get(1).push(4);
       graph.get(4).push(1);

       // 2
       graph.get(2).push(3);
       graph.get(3).push(2);

       // 3
       graph.get(4).push(5);
       graph.get(5).push(4);

       // 4
       graph.get(3).push(9);
       graph.get(9).push(3);

       // 5
       graph.get(6).push(9);
       graph.get(9).push(6);

       // 6
       graph.get(3).push(8);
       graph.get(8).push(3);

       // 7
       graph.get(10).push(4);
       graph.get(4).push(10);

       // 8
       graph.get(2).push(7);
       graph.get(7).push(2);

       // 9
       graph.get(3).push(6);
       graph.get(6).push(3);

       // 10
       graph.get(2).push(8);
       graph.get(8).push(2);

       // 11
       graph.get(9).push(2);
       graph.get(2).push(9);

       // 12
       graph.get(1).push(10);
       graph.get(10).push(1);

       // 13
       graph.get(9).push(10);
       graph.get(10).push(9);

       get(graph);

      // This code is contributed by Potta Lokesh
   </script>

Output

3

Time Complexity: O(N)
Auxiliary Space: O(N)