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[0], a[n-1]) and (a[1], a[n-2]) and so on, and their absolute difference sum will yield us the maximum-sum.
Implementation:
C++
// CPP program to find minimum and maximum // sum of absolute differences of pairs#include <bits/stdc++.h>using namespace std;// function to calculate minimum sumint 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 sumint 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 functionint main(){ int a[] = { 10, -10, 20, -40}; int n = sizeof(a) / sizeof(a[0]); 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 pairsimport 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 sumdef 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 sumdef 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 Codeif __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 pairsusing 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
<?php// PHP program to find minimum and maximum // sum of absolute differences of pairs// function to calculate minimum sumfunction calculate_min_sum($a, $n){ // sorts the array c++ stl sort($a); // initially min=0 and max=0 $min_sum = 0; // traverse to find the minimum sum for ($i = 1; $i < $n; $i += 2) { // the adjacent elements difference // will always be smaller $min_sum += abs($a[$i] - $a[$i - 1]); } return $min_sum;}// function to calculate maximum sumfunction calculate_max_sum($a, $n){ // sorts the array c++ stl sort($a); $max_sum = 0; // traverse to find the maximum sum for ($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 $max_sum;}// Driver Code$a = array(10, -10, 20, -40);$n = sizeof($a);echo("The minimum sum of pairs is " . calculate_min_sum($a, $n) . "\n");echo("The maximum sum of pairs is " . calculate_max_sum($a, $n));// This code is contributed by Ajit.?> |
Javascript
<script>// Javascript program to find minimum and maximum // sum of absolute differences of pairs// Function to calculate minimum sumfunction calculate_min_sum(a, n){ // Sorts the array c++ stl a.sort(); // Initially min=0 and max=0 let min_sum = 0; // Traverse to find the minimum sum for(let 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 sumfunction calculate_max_sum(a, n){ // Sorts the array c++ stl a.sort(); let max_sum = 0; // Traverse to find the maximum sum for(let i = 0; i < parseInt(n / 2, 10); i++) { // The farthest distant elements sum // will always be maximum max_sum += Math.abs(a[n - 1 - i] - a[i]); } return max_sum;}// Driver codelet a = [ 10, -10, 20, -40 ];let n = a.length; document.write("The minimum sum of pairs is " + calculate_min_sum(a, n) + "</br>"); document.write("The maximum sum of pairs is " + calculate_max_sum(a, n)); // This code is contributed by decode2207</script> |
Output
The minimum sum of pairs is 40 The maximum sum of pairs is 80
Time complexity : O(n log n)
This article is contributed by Raja Vikramaditya. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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