# Maximize count of nodes disconnected from all other nodes in a Graph

Given two integers **N** and **E** which denotes the number of nodes and the number of edges of an undirected graph, the task is to maximize the number of nodes which is not connected to any other node in the graph, without using any self-loops.

**Examples:**

Input:N = 5, E = 1Output:3Explanation:

Since there is only 1 edge in the graph which can be used to connect two nodes.

Therefore,threenode remains disconnected.

Input:N = 5, E = 2Output:2

**Approach:** The approach is based on the idea that to maximize the number of disconnected nodes, the new nodes will not be added to the graph until every two distinct nodes become connected. Below are the steps to solve this problem:

- Initialize two variables
**curr**and**rem**to store the nodes connected and the edges remaining unassigned respectively. - If
**rem**becomes 0, then the required answer will be**N – curr**. - Otherwise, increment the value of
**curr**by 1. - So, the maximum edges needed in the current step to keep every two distinct nodes connected is
**min(rem, curr)**. Subtract it from**rem**and increment**curr**. - Repeat this process until
**rem**reduces to zero. - Finally, print
**N – curr**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function which returns` `// the maximum number of` `// isolated nodes` `int` `maxDisconnected(` `int` `N, ` `int` `E)` `{` ` ` `// Used nodes` ` ` `int` `curr = 1;` ` ` `// Remaining edges` ` ` `int` `rem = E;` ` ` `// Count nodes used` ` ` `while` `(rem > 0) {` ` ` `rem = rem` ` ` `- min(` ` ` `curr, rem);` ` ` `curr++;` ` ` `}` ` ` `// If given edges are non-zero` ` ` `if` `(curr > 1) {` ` ` `return` `N - curr;` ` ` `}` ` ` `else` `{` ` ` `return` `N;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given N and E` ` ` `int` `N = 5, E = 1;` ` ` `// Function Call` ` ` `cout << maxDisconnected(N, E);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of` `// the above approach` `import` `java.util.*;` `class` `GFG{` `// Function which returns` `// the maximum number of` `// isolated nodes` `static` `int` `maxDisconnected(` `int` `N, ` `int` `E)` `{` ` ` ` ` `// Used nodes` ` ` `int` `curr = ` `1` `;` ` ` `// Remaining edges` ` ` `int` `rem = E;` ` ` `// Count nodes used` ` ` `while` `(rem > ` `0` `)` ` ` `{` ` ` `rem = rem - Math.min(` ` ` `curr, rem);` ` ` `curr++;` ` ` `}` ` ` `// If given edges are non-zero` ` ` `if` `(curr > ` `1` `)` ` ` `{` ` ` `return` `N - curr;` ` ` `}` ` ` `else` ` ` `{` ` ` `return` `N;` ` ` `}` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given N and E` ` ` `int` `N = ` `5` `, E = ` `1` `;` ` ` `// Function call` ` ` `System.out.print(maxDisconnected(N, E));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python3 implementation of` `# the above approach` `# Function which returns` `# the maximum number of` `# isolated nodes` `def` `maxDisconnected(N, E):` ` ` `# Used nodes` ` ` `curr ` `=` `1` ` ` `# Remaining edges` ` ` `rem ` `=` `E` ` ` `# Count nodes used` ` ` `while` `(rem > ` `0` `):` ` ` `rem ` `=` `rem ` `-` `min` `(curr, rem)` ` ` `curr ` `+` `=` `1` ` ` `# If given edges are non-zero` ` ` `if` `(curr > ` `1` `):` ` ` `return` `N ` `-` `curr` ` ` `else` `:` ` ` `return` `N` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# Given N and E` ` ` `N ` `=` `5` ` ` `E ` `=` `1` ` ` `# Function call` ` ` `print` `(maxDisconnected(N, E))` `# This code is contributed by mohit kumar 29` |

## C#

`// C# implementation of` `// the above approach` `using` `System;` `class` `GFG{` `// Function which returns` `// the maximum number of` `// isolated nodes` `static` `int` `maxDisconnected(` `int` `N,` ` ` `int` `E)` `{ ` ` ` `// Used nodes` ` ` `int` `curr = 1;` ` ` `// Remaining edges` ` ` `int` `rem = E;` ` ` `// Count nodes used` ` ` `while` `(rem > 0)` ` ` `{` ` ` `rem = rem - Math.Min(curr, rem);` ` ` `curr++;` ` ` `}` ` ` `// If given edges are non-zero` ` ` `if` `(curr > 1)` ` ` `{` ` ` `return` `N - curr;` ` ` `}` ` ` `else` ` ` `{` ` ` `return` `N;` ` ` `}` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `// Given N and E` ` ` `int` `N = 5, E = 1;` ` ` `// Function call` ` ` `Console.Write(maxDisconnected(N, E));` `}` `}` ` ` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// Javascript implementation of` `// the above approach` ` ` `// Function which returns` `// the maximum number of` `// isolated nodes` `function` `maxDisconnected(N,E)` `{` ` ` `// Used nodes` ` ` `let curr = 1;` ` ` ` ` `// Remaining edges` ` ` `let rem = E;` ` ` ` ` `// Count nodes used` ` ` `while` `(rem > 0)` ` ` `{` ` ` `rem = rem - Math.min(` ` ` `curr, rem);` ` ` `curr++;` ` ` `}` ` ` ` ` `// If given edges are non-zero` ` ` `if` `(curr > 1)` ` ` `{` ` ` `return` `N - curr;` ` ` `}` ` ` `else` ` ` `{` ` ` `return` `N;` ` ` `}` `}` `// Driver Code` `// Given N and E` `let N = 5, E = 1;` `// Function call` `document.write(maxDisconnected(N, E));` `// This code is contributed by unknown2108` `</script>` |

**Output:**

3

**Time Complexity:** O(E) **Auxiliary Space:** O(1)