最大化不属于图中任何边的节点数

给定一个具有 n 个节点和 m 个边的图。找到不属于任何边的最大可能节点数（m 将始终小于或等于完整图中的边数）。
例子：

Input: n = 3, m = 3
Output: Maximum Nodes Left Out: 0
Since it is a complete graph.

Input: n = 7, m = 6 
Output: Maximum Nodes Left Out: 3
We can construct a complete graph on 4 vertices using 6 edges.

方法：遍历所有 n 并查看如果我们制作一个完整的图，我们在哪些节点上获得的边数超过 m 说它是 K。答案是nk 。

可用于在 n 个节点上形成图的最大边数为n * (n – 1) / 2 （完整图）。
然后找到最大 n 的个数，它将使用 m 或小于 m 的边来形成一个完整的图。
如果仍然有边，那么它将只覆盖一个以上的节点，就好像它会覆盖一个以上的节点一样，这不是 n 的最大值。

以下是上述方法的实现：

C++

// C++ program to illustrate above approach
#include 
#define ll long long int
using namespace std;
 
// Function to return number of nodes left out
int answer(int n, int m)
{
    int i;
    for (i = 0; i <= n; i++) {
 
        // Condition to terminate, when
        // m edges are covered
        if ((i * (i - 1)) >= 2 * m)
            break;
    }
 
    return n - i;
}
 
// Driver Code
int main()
{
    int n = 7;
    int m = 6;
    cout << answer(n, m) << endl;
}

Java

// Java program to illustrate above approach
 
import java.io.*;
 
class GFG {
 
// Function to return number of nodes left out
static int answer(int n, int m)
{
    int i;
    for (i = 0; i <= n; i++) {
 
        // Condition to terminate, when
        // m edges are covered
        if ((i * (i - 1)) >= 2 * m)
            break;
    }
 
    return n - i;
}
 
        // Driver Code
    public static void main (String[] args) {
        int n = 7;
    int m = 6;
    System.out.print( answer(n, m));
    }
}
// This code is contributed by anuj_67..

Python3

# Python 3 program to illustrate
# above approach
 
# Function to return number of
# nodes left out
def answer(n, m):
    for i in range(0, n + 1, 1):
         
        # Condition to terminate, when
        # m edges are covered
        if ((i * (i - 1)) >= 2 * m):
            break
     
    return n - i
 
# Driver Code
if __name__ == '__main__':
    n = 7
    m = 6
    print(answer(n, m))
 
# This code is contributed
# by Surendra_Gangwar

C#

// C# program to illustrate
// above approach
using System;
 
class GFG
{
     
// Function to return number
// of nodes left out
static int answer(int n, int m)
{
    int i;
    for (i = 0; i <= n; i++)
    {
 
        // Condition to terminate, when
        // m edges are covered
        if ((i * (i - 1)) >= 2 * m)
            break;
    }
 
    return n - i;
}
 
// Driver Code
static public void Main ()
{
    int n = 7;
    int m = 6;
    Console.WriteLine(answer(n, m));
}
}
 
// This code is contributed
// by anuj_67

PHP

= 2 * $m)
            break;
    }
 
    return $n - $i;
}
 
// Driver Code
$n = 7;
$m = 6;
echo answer($n, $m) + "\n";
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

输出：