Minimum sum of absolute differences between pairs of a triplet from an array

Last Updated : 09 Jul, 2021

Given an array A[] consisting of positive integers, the task is to find the minimum value of |A[x] - A[y]| + |A[y] - A[z]| of any triplet (A[x], A[y], A[z]) from an array.

Examples:

Input: A[] = { 1, 1, 2, 3 }
Output: 1
Explanation:
For x = 0, y = 1, z = 2
|A[x] - A[y]| + |A[y] - A[z]| = 0 + 1 = 1, which is maximum possible

Input : A[] = { 1, 1, 1 }
Output : 0

Approach : The problem can be solved greedily. Follow the steps below to solve the problem:

Traverse the array.
Sort the array in ascending order.
Traverse the array using a variable i over indices [0, N - 3]. For every i^th index, set x = i, y = i + 1, z = i + 2
Calculate the sum of the triplet (x, y, z).
Update the minimum sum possible.
Print the minimum sum obtained.

Below is the implementation of the above approach:

C++

// C++ Program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find minimum
// sum of absolute differences
// of pairs of a triplet
int minimum_sum(int A[], int N)
{
    // Sort the array
    sort(A, A + N);

    // Stores the minimum sum
    int sum = INT_MAX;

    // Traverse the array
    for (int i = 0; i <= N - 3; i++) {

        // Update the minimum sum
        sum = min(sum,
                  abs(A[i] - A[i + 1]) +
                  abs(A[i + 1] - A[i + 2]));
    }

    // Print the minimum sum
    cout << sum;
}

// Driver Code
int main()
{

    // Input
    int A[] = { 1, 1, 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function call to find minimum 
    // sum of absolute differences
    // of pairs in a triplet
    minimum_sum(A, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
  
// Function to find minimum
// sum of absolute differences
// of pairs of a triplet
static int minimum_sum(int []A, int N)
{
  
    // Sort the array
    Arrays.sort(A);

    // Stores the minimum sum

    int sum = 2147483647;

    // Traverse the array
    for (int i = 0; i <= N - 3; i++) {

        // Update the minimum sum
        sum = Math.min(sum,Math.abs(A[i] - A[i + 1]) + Math.abs(A[i + 1] - A[i + 2]));
    }

    // Print the minimum sum
    return sum;
}

// Driver Code
public static void main(String[] args)
{
  
    // Input
    int []A = { 1, 1, 2, 3 };
    int N = A.length;

    // Function call to find minimum 
    // sum of absolute differences
    // of pairs in a triplet
    System.out.print(minimum_sum(A, N));
}
}

// This code is contributed by splevel62.

Python3

# Python 3 Program for the above approach
import sys

# Function to find minimum
# sum of absolute differences
# of pairs of a triplet
def minimum_sum(A, N):
  
    # Sort the array
    A.sort(reverse = False)

    # Stores the minimum sum
    sum = sys.maxsize

    # Traverse the array
    for i in range(N - 2):
      
        # Update the minimum sum
        sum = min(sum, abs(A[i] - A[i + 1]) + abs(A[i + 1] - A[i + 2]))

    # Print the minimum sum
    print(sum)

# Driver Code
if __name__ == '__main__':
  
    # Input
    A = [1, 1, 2, 3]
    N = len(A)

    # Function call to find minimum 
    # sum of absolute differences
    # of pairs in a triplet
    minimum_sum(A, N)
    
    # This code is contributed by ipg2016107

// C# Program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
   
// Function to find minimum
// sum of absolute differences
// of pairs of a triplet
static int minimum_sum(int []A, int N)
{
  
    // Sort the array
    Array.Sort(A);

    // Stores the minimum sum

    int sum = 2147483647;

    // Traverse the array
    for (int i = 0; i <= N - 3; i++) {

        // Update the minimum sum
        sum = Math.Min(sum,Math.Abs(A[i] - A[i + 1]) + Math.Abs(A[i + 1] - A[i + 2]));
    }

    // Print the minimum sum
    return sum;
}

// Driver Code
public static void Main()
{

    // Input
    int []A = { 1, 1, 2, 3 };
    int N = A.Length;

    // Function call to find minimum 
    // sum of absolute differences
    // of pairs in a triplet
    Console.WriteLine(minimum_sum(A, N));
}
}

// This code is contributed by bgangwar59.

JavaScript

<script>

// Javascript Program for the above approach

// Function to find minimum
// sum of absolute differences
// of pairs of a triplet
function minimum_sum( A, N)
{
    // Sort the array
    A.sort();

    // Stores the minimum sum
    var sum = 1000000000;

    // Traverse the array
    for (var i = 0; i <= N - 3; i++) {

        // Update the minimum sum
        sum = Math.min(sum,
                  Math.abs(A[i] - A[i + 1]) +
                  Math.abs(A[i + 1] - A[i + 2]));
    }

    // Print the minimum sum
    document.write(sum);
}

// Driver Code
// Input
var A = [ 1, 1, 2, 3 ];
var N = A.length;
// Function call to find minimum 
// sum of absolute differences
// of pairs in a triplet
minimum_sum(A, N);


</script>

Output:

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

Minimum sum of absolute difference of pairs of two arrays

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