Given ‘n’ vertices and ‘m’ edges of a graph. Find the minimum number and maximum number of isolated vertices that are possible in the graph.
Examples:
Input : 4 2 Output : Minimum 0 Maximum 1 1--2 3--4 <---Minimum - No isolated vertex 1--2 <--- Maximum - 1 Isolated vertex i.e. 4 | 3 Input : 5 2 Output : Minimum 1 Maximum 2 1--2 3--4 5 <-- Minimum - 1 isolated vertex i.e. 5 1--2 4 5 <-- Maximum - 2 isolated vertex i.e. 4 and 5 | 3
- For minimum number of isolated vertices, we connect two vertices by only one edge. Each vertex should be only connected to one other vertex and each vertex should have degree one
Thus if the number of edges is ‘m’, and if ‘n’ vertices <=2 * ‘m’ edges, there is no isolated vertex and if this condition is false, there are n-2*m isolated vertices. - For maximum number of isolated vertices, we create a polygon such that each vertex is connected to other vertex and each vertex has a diagonal with every other vertex. Thus, number of diagonals from one vertex to other vertex of n sided polygon is n*(n-3)/2 and number of edges connecting adjacent vertices is n. Thus, total number of edges is n*(n-1)/2.
Below is the implementation of above approach.
C++
// CPP program to find maximum/minimum number // of isolated vertices. #include <bits/stdc++.h> using namespace std;
// Function to find out the minimum and // maximum number of isolated vertices void find( int n, int m)
{ // Condition to find out minimum number
// of isolated vertices
if (n <= 2 * m)
cout << "Minimum " << 0 << endl;
else
cout << "Minimum " << n - 2 * m << endl;
// To find out maximum number of isolated
// vertices
// Loop to find out value of number of
// vertices that are connected
int i;
for (i = 0; i <= n; i++) {
if (i * (i - 1) / 2 >= m)
break ;
}
cout << "Maximum " << n - i;
} // Driver Function int main()
{ // Number of vertices
int n = 4;
// Number of edges
int m = 2;
// Calling the function to maximum and
// minimum number of isolated vertices
find(n, m);
return 0;
} |
Java
// Java program to find maximum/minimum number // of isolated vertices. import java.io.*;
class GFG {
// Function to find out the minimum and // maximum number of isolated vertices static void find( int n, int m)
{ // Condition to find out minimum number
// of isolated vertices
if (n <= 2 * m)
System.out.println( "Minimum " + 0 );
else
System.out.println( "Minimum " + (n - 2 * m));
// To find out maximum number of isolated
// vertices
// Loop to find out value of number of
// vertices that are connected
int i;
for (i = 0 ; i <= n; i++) {
if (i * (i - 1 ) / 2 >= m)
break ;
}
System.out.println( "Maximum " + (n - i));
} // Driver Function public static void main (String[] args) {
// Number of vertices
int n = 4 ;
// Number of edges
int m = 2 ;
// Calling the function to maximum and
// minimum number of isolated vertices
find(n, m);
}
} //This code is contributed by inder_verma. |
Python3
# Python3 program to find maximum/minimum # number of isolated vertices. # Function to find out the minimum and # maximum number of isolated vertices def find(n, m) :
# Condition to find out minimum
# number of isolated vertices
if (n < = 2 * m):
print ( "Minimum " , 0 )
else :
print ( "Minimum " , n - 2 * m )
# To find out maximum number of
# isolated vertices
# Loop to find out value of number
# of vertices that are connected
for i in range ( 0 , n + 1 ):
if (i * (i - 1 ) / / 2 > = m):
break
print ( "Maximum " , n - i)
# Driver Code if __name__ = = '__main__' :
# Number of vertices
n = 4
# Number of edges
m = 2
# Calling the function to maximum and
# minimum number of isolated vertices
find(n, m)
# This code is contributed by # SHUBHAMSINGH10 |
C#
// C# program to find maximum/ // minimum number of isolated vertices. using System;
class GFG
{ // Function to find out the // minimum and maximum number // of isolated vertices static void find( int n, int m)
{ // Condition to find out minimum
// number of isolated vertices
if (n <= 2 * m)
Console.WriteLine( "Minimum " + 0);
else
Console.WriteLine( "Minimum " +
(n - 2 * m));
// To find out maximum number
// of isolated vertices
// Loop to find out value of
// number of vertices that
// are connected
int i;
for (i = 0; i <= n; i++)
{
if (i * (i - 1) / 2 >= m)
break ;
}
Console.WriteLine( "Maximum " + (n - i));
} // Driver Code public static void Main ()
{ // Number of vertices
int n = 4;
// Number of edges
int m = 2;
// Calling the function to
// maximum and minimum number
// of isolated vertices
find(n, m);
} } // This code is contributed // by inder_verma. |
PHP
<?php // PHP program to find maximum/minimum // number of isolated vertices. // Function to find out the // minimum and maximum number // of isolated vertices function find( $n , $m )
{ // Condition to find out minimum
// number of isolated vertices
if ( $n <= 2 * $m )
echo "Minimum 0\n" ;
else
echo "Minimum " , ( $n - 2 * $m );
// To find out maximum number
// of isolated vertices
// Loop to find out value of number
// of vertices that are connected
for ( $i = 0; $i <= $n ; $i ++)
{
if ( $i * ( $i - 1) / 2 >= $m )
break ;
}
echo "Maximum " , ( $n - $i );
} // Driver Code // Number of vertices $n = 4;
// Number of edges $m = 2;
// Calling the function to // maximum and minimum number // of isolated vertices find( $n , $m );
// This code is contributed // by inder_verma ?> |
Javascript
<script> // Javascript program to find maximum/
// minimum number of isolated vertices.
// Function to find out the
// minimum and maximum number
// of isolated vertices
function find(n, m)
{
// Condition to find out minimum
// number of isolated vertices
if (n <= 2 * m)
document.write( "Minimum " + 0 + "</br>" );
else
document.write( "Minimum " + (n - 2 * m) + "</br>" );
// To find out maximum number
// of isolated vertices
// Loop to find out value of
// number of vertices that
// are connected
let i;
for (i = 0; i <= n; i++)
{
if (i * parseInt((i - 1) / 2, 10) >= m)
break ;
}
document.write( "Maximum " + (n - i));
}
// Number of vertices
let n = 4;
// Number of edges
let m = 2;
// Calling the function to
// maximum and minimum number
// of isolated vertices
find(n, m);
// This code is contributed by divyeshrabadiya07. </script> |
Output
Minimum 0 Maximum 1
Time Complexity: O(n)
Auxiliary Space: O(1)
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