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Minimum and Maximum sum of absolute differences of pairs
• Difficulty Level : Easy
• Last Updated : 30 Apr, 2021

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 ```

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 */`

## Python3

 `# 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

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## Javascript

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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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