# Minimum and Maximum sum of absolute differences of pairs

Given an array of N integers where N is even, find the minimum and maximum sum of absolute difference of N/2 pairs formed by pairing every element with one other element.

Examples:

```Input: a[] = {10, -10, 20, -40}
Output: min_sum = 40, max_sum = 80
Explanation: Pairs selected for minimum sum
(-10, -40) and (10, 20)
min_sum = |-10 - -40| + |20 - 10| = 40
Pairs selected for maximum sum
(-10, 20) and (-40, 10)
max_sum = |-10 - 20| + |10 - -40| = 80

Input: a[] = {20, -10, -1, 30}
Output: min_sum = 19, max_sum = 61
Explanation: Pairs selected for minimum sum
(-1, -10) and (20, 30)
min_sum = |-1 - -10| + |20 - 30| = 19
Pairs selected for maximum sum
(-1, 30) and (-10, 20)
max_sum = |-1 - 30| + |-10 - 20| = 61
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The most common observation will be that for minimum sum of differences we need the closest elements together as a pair and for the maximum sum we need the farthest elements together as a pair. So, we can simply sort the given list of elements and the closest pairs will be a[i], a[i+1], their absolute difference sum will yield us the minimum sum. The farthest will be (a, a[n-1]) and (a, a[n-2]) and so on, and their absolute difference sum will yield us the maximum-sum.

## C++

 `// CPP program to find minimum and maximum  ` `// sum of absolute differences of pairs ` `#include ` `using` `namespace` `std; ` ` `  `// function to calculate minimum sum ` `int` `calculate_min_sum(``int` `a[], ``int` `n) ` `{ ` `    ``// sorts the array c++ stl ` `    ``sort(a, a + n); ` ` `  `    ``// initially min=0 and max=0 ` `    ``int` `min_sum = 0; ` ` `  `    ``// traverse to find the minimum sum ` `    ``for` `(``int` `i = 1; i < n; i += 2) { ` ` `  `        ``// the adjacent elements difference ` `        ``// will always be smaller ` `        ``min_sum += ``abs``(a[i] - a[i - 1]); ` `    ``} ` `    ``return` `min_sum; ` `} ` ` `  `// function to calculate maximum sum ` `int` `calculate_max_sum(``int` `a[], ``int` `n) ` `{ ` `    ``// sorts the array c++ stl ` `    ``sort(a, a + n); ` ` `  `    ``int` `max_sum = 0; ` ` `  `    ``// traverse to find the maximum sum ` `    ``for` `(``int` `i = 0; i < n / 2; i++) { ` `         `  `        ``// the farthest distant elements sum  ` `        ``// will always be maximum ` `        ``max_sum += ``abs``(a[n - 1 - i] - a[i]); ` `    ``} ` `    ``return` `max_sum; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `a[] = { 10, -10, 20, -40}; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``cout << ``"The minimum sum of pairs is "`  `         ``<< calculate_min_sum(a, n) << endl; ` ` `  `    ``cout << ``"The maximum sum of pairs is "`  `         ``<< calculate_max_sum(a, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find minimum and maximum  ` `// sum of absolute differences of pairs ` `import` `java.util.Arrays; ` ` `  `class` `GFG ` `{ ` `    ``// function to calculate minimum sum ` `    ``static` `int` `calculate_min_sum(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// sorts the array c++ stl ` `        ``Arrays.sort(a); ` ` `  `        ``// initially min=0 and max=0 ` `        ``int` `min_sum = ``0``; ` ` `  `        ``// traverse to find the minimum sum ` `        ``for` `(``int` `i = ``1``; i < n; i += ``2``) { ` ` `  `            ``// the adjacent elements difference ` `            ``// will always be smaller ` `            ``min_sum += Math.abs(a[i] - a[i - ``1``]); ` `        ``} ` `        ``return` `min_sum; ` `    ``} ` ` `  `    ``// function to calculate maximum sum ` `    ``static` `int` `calculate_max_sum(``int``[] a, ``int` `n) ` `    ``{ ` `        ``// sorts the array c++ stl ` `        ``Arrays.sort(a); ` ` `  `        ``int` `max_sum = ``0``; ` ` `  `        ``// traverse to find the maximum sum ` `        ``for` `(``int` `i = ``0``; i < n / ``2``; i++) { ` `         `  `            ``// the farthest distant elements sum  ` `            ``// will always be maximum ` `            ``max_sum += Math.abs(a[n - ``1` `- i] - a[i]); ` `        ``} ` `        ``return` `max_sum; ` `    ``} ` ` `  `    ``// Driver program to test above function     ` `    ``public` `static` `void` `main (String[] args) { ` `    ``int``[] a = { ``10``, -``10``, ``20``, -``40``}; ` ` `  `    ``int` `n = a.length; ` `     `  `    ``System.out.println(``"The minimum sum of pairs is "` `+ ` `                          ``calculate_min_sum(a, n));  ` ` `  `    ``System.out.println(``"The maximum sum of pairs is "` `+ ` `                           ``calculate_max_sum(a, n));  ` `     `  `    ``} ` `} ` ` `  `/* This code is contributed by Mr. Somesh Awasthi */`

## Python 3

 `# Python 3 program to find minimum and maximum  ` `# sum of absolute differences of pairs ` ` `  `# function to calculate minimum sum ` `def` `calculate_min_sum( a, n): ` ` `  `    ``# sorts the array c++ stl ` `    ``a.sort() ` ` `  `    ``# initially min=0 and max=0 ` `    ``min_sum ``=` `0` ` `  `    ``# traverse to find the minimum sum ` `    ``for` `i ``in` `range``(``1``, n, ``2``): ` ` `  `        ``# the adjacent elements difference ` `        ``# will always be smaller ` `        ``min_sum ``+``=` `abs``(a[i] ``-` `a[i ``-` `1``]) ` `     `  `    ``return` `min_sum ` ` `  `# function to calculate maximum sum ` `def` `calculate_max_sum(a, n): ` ` `  `    ``# sorts the array c++ stl ` `    ``a.sort() ` ` `  `    ``max_sum ``=` `0` ` `  `    ``# traverse to find the maximum sum ` `    ``for` `i ``in` `range``(n ``/``/` `2``): ` `         `  `        ``# the farthest distant elements sum  ` `        ``max_sum ``+``=` `abs``(a[n ``-` `1` `-` `i] ``-` `a[i]) ` `    ``return` `max_sum ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``a ``=` `[ ``10``, ``-``10``, ``20``, ``-``40``] ` ` `  `    ``n ``=` `len``(a) ` ` `  `    ``print``(``"The minimum sum of pairs is"``,  ` `                ``calculate_min_sum(a, n)) ` ` `  `    ``print``( ``"The maximum sum of pairs is"``,  ` `                 ``calculate_max_sum(a, n)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# program to find minimum and maximum  ` `// sum of absolute differences of pairs ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// function to calculate minimum sum ` `    ``static` `int` `calculate_min_sum(``int` `[]a, ``int` `n) ` `    ``{ ` `        ``// sorts the array c++ stl ` `        ``Array.Sort(a); ` ` `  `        ``// initially min=0 and max=0 ` `        ``int` `min_sum = 0; ` ` `  `        ``// traverse to find the minimum sum ` `        ``for` `(``int` `i = 1; i < n; i += 2) { ` ` `  `            ``// the adjacent elements difference ` `            ``// will always be smaller ` `            ``min_sum += Math.Abs(a[i] - a[i - 1]); ` `        ``} ` `        ``return` `min_sum; ` `    ``} ` ` `  `    ``// Function to calculate maximum sum ` `    ``static` `int` `calculate_max_sum(``int` `[]a, ``int` `n) ` `    ``{ ` `        ``// sorts the array c++ stl ` `        ``Array.Sort(a); ` ` `  `        ``int` `max_sum = 0; ` ` `  `        ``// Traverse to find the maximum sum ` `        ``for` `(``int` `i = 0; i < n / 2; i++) { ` `         `  `            ``// the farthest distant elements sum  ` `            ``// will always be maximum ` `            ``max_sum += Math.Abs(a[n - 1 - i] - a[i]); ` `        ``} ` `        ``return` `max_sum; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `    ``int` `[]a = { 10, -10, 20, -40}; ` ` `  `    ``int` `n = a.Length; ` `     `  `    ``Console.WriteLine(``"The minimum sum of pairs is "` `+ ` `                            ``calculate_min_sum(a, n));  ` ` `  `    ``Console.Write(``"The maximum sum of pairs is "` `+ ` `                         ``calculate_max_sum(a, n));  ` `     `  `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

Output:

```The minimum sum of pairs is 40
The maximum sum of pairs is 80
```

Time complexity : O(n log n)

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