图中的最大和最小孤立顶点

给定图的“ n”个顶点和“ m”个边。找到图形中可能的孤立顶点的最小数量和最大数量。
例子：

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

1. 对于最少数量的孤立顶点，我们仅通过一条边连接两个顶点。每个顶点应仅连接到另一个顶点，并且每个顶点应具有一级
   因此，如果边的数量为’m’，并且如果’n’个顶点<= 2 *’m’个边，则不存在孤立的顶点，并且如果此条件为false，则存在n-2 * m个孤立的顶点。
2. 为了获得最大数量的孤立顶点，我们创建了一个多边形，使每个顶点都连接到另一个顶点，并且每个顶点与其他每个顶点都有一个对角线。因此，从n个面的多边形的一个顶点到另一顶点的对角线的数量为n *(n-3)/ 2，并且连接相邻顶点的边的数量为n。因此，边缘的总数为n *(n-1)/ 2。

下面是上述方法的实现。

C++

// CPP program to find maximum/minimum number
// of isolated vertices.
#include 
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 = 1; 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 = 1; 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(1, 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 = 1; 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

= $m)
            break;
    }
    echo "Maximum " , ($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
?>

输出：

Minimum 0
Maximum 1

时间复杂度– O(n)