Maximize number of nodes which are not part of any edge in a Graph

Given a graph with n nodes and m edges. Find the maximum possible number of nodes which are not part of any edge (m will always be less than or equal to a number of edges in complete graph).

Examples:

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.

Approach: Iterate over all n and see at which a number of nodes if we make a complete graph we obtain a number of edges more than m say it is K. Answer is n-k. 

• Maximum number of edges which can be used to form a graph on n nodes is n * (n – 1) / 2 (A complete Graph).
• Then find number of maximum n, which will use m or less than m edges to form a complete graph.
• If still edges are left, then it will cover only one more node, as if it would have covered more than one node than, this is not the maximum value of n.

Below is the implementation of above approach: 

C++

// C++ program to illustrate above approach
#include <bits/stdc++.h>
#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

<?php
// PHP program to illustrate
// above approach

// Function to return number
// of nodes left out
function answer($n, $m)
{
    for ($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
$n = 7;
$m = 6;
echo answer($n, $m) + "\n";

// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

<script>

// Javascript implementation of the above approach

// Function to return number of nodes left out
function answer(n, m)
{
    let 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 program

    let n = 7;
    let m = 6;
    document.write( answer(n, m));

</script>

Output

3

Complexity Analysis:

• Time Complexity: O(n) where n is the number of nodes.
• Auxiliary Space: O(1)