# 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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.