Given an of integers of size N. The task is to separate these integers into two groups g1 and g2 such that (sum of elements of g1) – (sum of elements of g2) becomes maximum. Your task is to print the value of result. We may keep one subset as empty.**Examples:**

Input :3, 7, -4, 10, -11, 2Output :37

Explanation:

g1: 3, 7, 10, 2

g2: -4, -11

result = ( 3 + 7 + 10 + 2 ) – ( -4 + -11) = 22 – (-15) = 37Input :2, 2, -2, -2Output :8

The idea is to group integers according to their sign value i.e., we group positive integers as g1 and negative integers as g2.

Since, – ( -g2 ) = +g2

Therefore, result becomes g1 + |g2|.

## C++

`// CPP program to make two subsets with` `// maximum difference.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `maxDiff(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `// We move all negative elements into` ` ` `// one set. So we add negation of negative` ` ` `// numbers to maximize difference` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `sum = sum + ` `abs` `(arr[i]);` ` ` ` ` `return` `sum;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 3, 7, -4, 10, -11, 2 };` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << maxDiff(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program to make two subsets with` `// maximum difference.` `import` `java.util.*;` `class` `solution` `{` `static` `int` `maxDiff(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `sum = ` `0` `;` ` ` `// We move all negative elements into` ` ` `// one set. So we add negation of negative` ` ` `// numbers to maximize difference` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `sum = sum + Math.abs(arr[i]);` ` ` ` ` `return` `sum;` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `[]arr = { ` `3` `, ` `7` `, -` `4` `, ` `10` `, -` `11` `, ` `2` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(maxDiff(arr, n));` `}` `}` |

## Python3

`# Python3 program to make two subsets` `# with maximum difference.` `def` `maxDiff(arr, n) :` ` ` `sum` `=` `0` ` ` `# We move all negative elements into` ` ` `# one set. So we add negation of negative` ` ` `# numbers to maximize difference` ` ` `for` `i ` `in` `range` `(n) :` ` ` `sum` `+` `=` `abs` `(arr[i])` ` ` ` ` `return` `sum` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `3` `, ` `7` `, ` `-` `4` `, ` `10` `, ` `-` `11` `, ` `2` `]` ` ` `n ` `=` `len` `(arr)` ` ` `print` `(maxDiff(arr, n))` `# This code is contributed by Ryuga` |

## C#

`using` `System;` `// C# program to make two subsets with` `// maximum difference.` `public` `class` `solution` `{` `public` `static` `int` `maxDiff(` `int` `[] arr, ` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `// We move all negative elements into` ` ` `// one set. So we add negation of negative` ` ` `// numbers to maximize difference ` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `sum = sum + Math.Abs(arr[i]);` ` ` `}` ` ` `return` `sum;` `}` `// Driver Code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `int` `[] arr = ` `new` `int` `[] {3, 7, -4, 10, -11, 2};` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(maxDiff(arr, n));` `}` `}` ` ` `// This code is contributed by Shrikant13` |

## PHP

`<?php` `// PHP program to make two subsets` `// with maximum difference.` `function` `maxDiff(` `$arr` `, ` `$n` `)` `{` ` ` `$sum` `= 0;` ` ` `// We move all negative elements` ` ` `// into one set. So we add negation` ` ` `// of negative numbers to maximize` ` ` `// difference` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `$sum` `= ` `$sum` `+ ` `abs` `(` `$arr` `[` `$i` `]);` ` ` ` ` `return` `$sum` `;` `}` `// Driver Code` `$arr` `= ` `array` `( 3, 7, -4, 10, -11, 2 );` `$n` `= sizeof(` `$arr` `);` `echo` `maxDiff(` `$arr` `, ` `$n` `);` ` ` `// This code is contributed by Sachin.` `?>` |

## Javascript

`<script>` `// javascript program to make two subsets with` `// maximum difference.` `function` `maxDiff(arr , n)` `{` ` ` `var` `sum = 0;` ` ` `// We move all negative elements into` ` ` `// one set. So we add negation of negative` ` ` `// numbers to maximize difference` ` ` `for` `(i = 0; i < n; i++)` ` ` `sum = sum + Math.abs(arr[i]);` ` ` ` ` `return` `sum;` `}` `// Driver Code` `var` `arr = [ 3, 7, -4, 10, -11, 2 ];` `var` `n = arr.length;` `document.write(maxDiff(arr, n));` `// This code is contributed by Princi Singh` `</script>` |

**Output:**

37

**Time Complexity: **O(n)

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