Given an undirected graph having N nodes, the task is to print the nodes having minimum and maximum degree.
Examples:
Input: 1-----2 | | 3-----4 Output: Nodes with maximum degree : 1 2 3 4 Nodes with minimum degree : 1 2 3 4 Every node has a degree of 2. Input: 1 / \ 2 3 / 4 Output: Nodes with maximum degree : 1 2 Nodes with minimum degree : 3 4
Approach: For an undirected graph, the degree of a node is the number of edges incident to it, so the degree of each node can be calculated by counting its frequency in the list of edges. Hence the approach is to use a map to calculate the frequency of every vertex from the edge list and use the map to find the nodes having maximum and minimum degrees.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to print the nodes having // maximum and minimum degree void minMax( int edges[][2], int len, int n)
{ // Map to store the degrees of every node
map< int , int > m;
for ( int i = 0; i < len; i++) {
// Storing the degree for each node
m[edges[i][0]]++;
m[edges[i][1]]++;
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0;
int mini = n;
for ( int i = 1; i <= n; i++) {
maxi = max(maxi, m[i]);
mini = min(mini, m[i]);
}
// Printing all the nodes with maximum degree
cout << "Nodes with maximum degree : " ;
for ( int i = 1; i <= n; i++) {
if (m[i] == maxi)
cout << i << " " ;
}
cout << endl;
// Printing all the nodes with minimum degree
cout << "Nodes with minimum degree : " ;
for ( int i = 1; i <= n; i++) {
if (m[i] == mini)
cout << i << " " ;
}
} // Driver code int main()
{ // Count of nodes and edges
int n = 4, m = 6;
// The edge list
int edges[][2] = { { 1, 2 },
{ 1, 3 },
{ 1, 4 },
{ 2, 3 },
{ 2, 4 },
{ 3, 4 } };
minMax(edges, m, 4);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to print the nodes having // maximum and minimum degree static void minMax( int edges[][], int len, int n)
{ // Map to store the degrees of every node
HashMap<Integer,
Integer> m = new HashMap<Integer,
Integer>();
for ( int i = 0 ; i < len; i++)
{
// Storing the degree for each node
if (m.containsKey(edges[i][ 0 ]))
{
m.put(edges[i][ 0 ], m.get(edges[i][ 0 ]) + 1 );
}
else
{
m.put(edges[i][ 0 ], 1 );
}
if (m.containsKey(edges[i][ 1 ]))
{
m.put(edges[i][ 1 ], m.get(edges[i][ 1 ]) + 1 );
}
else
{
m.put(edges[i][ 1 ], 1 );
}
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0 ;
int mini = n;
for ( int i = 1 ; i <= n; i++)
{
maxi = Math.max(maxi, m.get(i));
mini = Math.min(mini, m.get(i));
}
// Printing all the nodes with maximum degree
System.out.print( "Nodes with maximum degree : " );
for ( int i = 1 ; i <= n; i++)
{
if (m.get(i) == maxi)
System.out.print(i + " " );
}
System.out.println();
// Printing all the nodes with minimum degree
System.out.print( "Nodes with minimum degree : " );
for ( int i = 1 ; i <= n; i++)
{
if (m.get(i) == mini)
System.out.print(i + " " );
}
} // Driver code public static void main(String[] args)
{ // Count of nodes and edges
int n = 4 , m = 6 ;
// The edge list
int edges[][] = {{ 1 , 2 }, { 1 , 3 },
{ 1 , 4 }, { 2 , 3 },
{ 2 , 4 }, { 3 , 4 }};
minMax(edges, m, 4 );
} } // This code is contributed by PrinciRaj1992 |
# Python3 implementation of the approach # Function to print the nodes having # maximum and minimum degree def minMax(edges, leng, n) :
# Map to store the degrees of every node
m = {};
for i in range (leng) :
m[edges[i][ 0 ]] = 0 ;
m[edges[i][ 1 ]] = 0 ;
for i in range (leng) :
# Storing the degree for each node
m[edges[i][ 0 ]] + = 1 ;
m[edges[i][ 1 ]] + = 1 ;
# maxi and mini variables to store
# the maximum and minimum degree
maxi = 0 ;
mini = n;
for i in range ( 1 , n + 1 ) :
maxi = max (maxi, m[i]);
mini = min (mini, m[i]);
# Printing all the nodes
# with maximum degree
print ( "Nodes with maximum degree : " ,
end = "")
for i in range ( 1 , n + 1 ) :
if (m[i] = = maxi) :
print (i, end = " " );
print ()
# Printing all the nodes
# with minimum degree
print ( "Nodes with minimum degree : " ,
end = "")
for i in range ( 1 , n + 1 ) :
if (m[i] = = mini) :
print (i, end = " " );
# Driver code if __name__ = = "__main__" :
# Count of nodes and edges
n = 4 ; m = 6 ;
# The edge list
edges = [[ 1 , 2 ], [ 1 , 3 ],
[ 1 , 4 ], [ 2 , 3 ],
[ 2 , 4 ], [ 3 , 4 ]];
minMax(edges, m, 4 );
# This code is contributed by AnkitRai01 |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ // Function to print the nodes having // maximum and minimum degree static void minMax( int [,]edges, int len, int n)
{ // Map to store the degrees of every node
Dictionary< int , int > m = new Dictionary< int , int >();
for ( int i = 0; i < len; i++)
{
// Storing the degree for each node
if (m.ContainsKey(edges[i, 0]))
{
m[edges[i, 0]] = m[edges[i, 0]] + 1;
}
else
{
m.Add(edges[i, 0], 1);
}
if (m.ContainsKey(edges[i, 1]))
{
m[edges[i, 1]] = m[edges[i, 1]] + 1;
}
else
{
m.Add(edges[i, 1], 1);
}
}
// maxi and mini variables to store
// the maximum and minimum degree
int maxi = 0;
int mini = n;
for ( int i = 1; i <= n; i++)
{
maxi = Math.Max(maxi, m[i]);
mini = Math.Min(mini, m[i]);
}
// Printing all the nodes with maximum degree
Console.Write( "Nodes with maximum degree : " );
for ( int i = 1; i <= n; i++)
{
if (m[i] == maxi)
Console.Write(i + " " );
}
Console.WriteLine();
// Printing all the nodes with minimum degree
Console.Write( "Nodes with minimum degree : " );
for ( int i = 1; i <= n; i++)
{
if (m[i] == mini)
Console.Write(i + " " );
}
} // Driver code public static void Main(String[] args)
{ // Count of nodes and edges
int m = 6;
// The edge list
int [,]edges = {{ 1, 2 }, { 1, 3 },
{ 1, 4 }, { 2, 3 },
{ 2, 4 }, { 3, 4 }};
minMax(edges, m, 4);
} } // This code is contributed by 29AjayKumar |
// JavaScript implementation of the approach // Function to print the nodes having // maximum and minimum degree function minMax(edges, len, n) {
// Map to store the degrees of every node
let m = new Map();
for (let i = 0; i < len; i++) {
// Storing the degree for each node
if (m.has(edges[i][0])) {
m.set(edges[i][0], m.get(edges[i][0]) + 1);
} else {
m.set(edges[i][0], 1);
}
if (m.has(edges[i][1])) {
m.set(edges[i][1], m.get(edges[i][1]) + 1);
} else {
m.set(edges[i][1], 1);
}
}
// maxi and mini variables to store
// the maximum and minimum degree
let maxi = 0;
let mini = n;
for (let i = 1; i <= n; i++) {
maxi = Math.max(maxi, m.get(i));
mini = Math.min(mini, m.get(i));
}
// Printing all the nodes with maximum degree
console.log( "Nodes with maximum degree : " , end = " " );
for (let i = 1; i <= n; i++) {
if (m.get(i) === maxi) {
console.log(i + " " );
}
}
console.log();
// Printing all the nodes with minimum degree
console.log( "Nodes with minimum degree : " , end = " " );
for (let i = 1; i <= n; i++) {
if (m.get(i) === mini) {
console.log(i + " " );
}
}
} // Driver code let n = 4, m = 6; // The edge list let edges = [ [1, 2],
[1, 3],
[1, 4],
[2, 3],
[2, 4],
[3, 4]
]; minMax(edges, m, 4); |
Nodes with maximum degree : 1 2 3 4 Nodes with minimum degree : 1 2 3 4
Time Complexity: O(M*logN + N).
Auxiliary Space: O(N).