Maximum sum of absolute differences between distinct pairs of a triplet from an array

• Difficulty Level : Basic
• Last Updated : 06 Apr, 2021

Given an array arr[] consisting of N integers, the task is to find the maximum sum of absolute differences between all distinct pairs of the triplet in the array.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 6
Explanation:
The valid triplet is (1, 3, 4) as sum = |1 – 4| + |1 – 3| + |3 – 4| = 6, which is the maximum among all the triplets.

Input: arr[] = {2, 2, 2}
Output: 0

Approach: The idea is to solve the given problem is to sort the array in ascending order and find the sum of absolute differences between the pairs of the first and the last two elements of the array. Follow the steps below to solve the problem:

• Initialize a variable, say sum, to store the maximum possible sum.
• Sort the given array arr[] in ascending order.
• Find the sum of differences between the pairs of the first and the last two elements of the array i.e., sum = (arr[N – 2] – arr) + (arr[N – 1] – arr) + (arr[N – 2] – arr[N – 1]).
• After completing the above steps, print the value of the sum as the result.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to find the maximum sum of// absolute differences between// distinct pairs of triplet in arrayvoid maximumSum(int arr[], int N){    // Stores the maximum sum    int sum;     // Sort the array in    // ascending order    sort(arr, arr + N);     // Sum of differences between    // pairs of the triplet    sum = (arr[N - 1] - arr)          + (arr[N - 2] - arr)          + (arr[N - 1] - arr[N - 2]);     // Print the sum    cout << sum;} // Driver Codeint main(){    int arr[] = { 1, 3, 4, 2 };    int N = sizeof(arr) / sizeof(arr);    maximumSum(arr, N);     return 0;}

Java

 // Java program for the above approachimport java.util.*;class GFG{   // Function to find the maximum sum of  // absolute differences between  // distinct pairs of triplet in array  static void maximumSum(int[] arr, int N)  {     // Stores the maximum sum    int sum;     // Sort the array in    // ascending order    Arrays.sort(arr);     // Sum of differences between    // pairs of the triplet    sum = (arr[N - 1] - arr) + (arr[N - 2] - arr)      + (arr[N - 1] - arr[N - 2]);     // Print the sum    System.out.println(sum);  }   // Driver Code  public static void main(String[] args)  {    int[] arr = { 1, 3, 4, 2 };    int N = arr.length;    maximumSum(arr, N);  }} // This code is contributed by susmitakundugoaldanga.

Python3

 # Python program for the above approach # Function to find the maximum sum of# absolute differences between# distinct pairs of triplet in arraydef maximumSum(arr, N):       # Stores the maximum sum    sum = 0         # Sort the array in    # ascending order    arr.sort()         # Sum of differences between    # pairs of the triplet    sum = (arr[N - 1] - arr) + (arr[N - 2] - arr) + (arr[N - 1] - arr[N - 2]);     # Print the sum    print(sum) # Driver Codearr = [ 1, 3, 4, 2 ]N = len(arr)maximumSum(arr, N) # This code is contributed by rohitsingh07052.

C#

 // C# program for the above approachusing System;using System.Collections.Generic; class GFG{     // Function to find the maximum sum of    // absolute differences between    // distinct pairs of triplet in array    static void maximumSum(int[] arr, int N)    {               // Stores the maximum sum        int sum;         // Sort the array in        // ascending order        Array.Sort(arr);         // Sum of differences between        // pairs of the triplet        sum = (arr[N - 1] - arr) + (arr[N - 2] - arr)              + (arr[N - 1] - arr[N - 2]);         // Print the sum        Console.Write(sum);    }     // Driver Code    public static void Main()    {        int[] arr = { 1, 3, 4, 2 };        int N = arr.Length;        maximumSum(arr, N);    }} // This code is contributed by chitranayal.

Javascript


Output:
6

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

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