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Finding the Sum of Minimum Absolute Difference of the given Array

Program to find the sum of minimum absolute difference of the array is discussed here. An array of distinct elements is given as input and the sum of minimum absolute difference of each array element has to be found. The minimum absolute difference is calculated using the formula,

Minimum Absolute Difference (a) = min(abs(a arr[j])) ; where 1 <= j <= n and j != i, abs is the absolute value.

For example, consider the following array given as input : arr = {1, 3, 9, 3, 6}

The optimal solution is to choose x = 3, which produces the sum

|1 3| + |3 3| + |9 3| + |3 3| + |6 3| = 2 + 0 + 6 + 0 + 3 = 11

Output : 11

Algorithm

The given input array is sorted.

For the first element of the array, its minimum absolute difference is calculated using the second array element.

For the last array element, its minimum absolute difference is calculated using the second last array element.

For the other array elements, minimum absolute difference for an element at index is calculated as follow:

minAbsDiff= min( abs(arr[i] arr[i-1]), abs(ar[i] arr[i+1]) ).

Program to find the sum of minimum absolute difference of the given array

C

// C program to find the sum of minimum absolute difference of the array

#include

#include

int min(int a, int b)

{

if(a>b)

return b;

else

r/eturn a;

}

// Function to find absolute sum

int abs_sum(int arr[], int n)

{

int sum = 0;

sum += abs(arr[0] – arr[1]);

sum += abs(arr[n-1] – arr[n-2]);

for (int i=1; i

sum += min(abs(arr[i] – arr[i-1]), abs(arr[i] – arr[i+1]));  // Total sum of absolute difference

return sum;

}

// Function to sort the elements

void sort(int a[], int n)

{

for(int i = 0; i < n>

{

for(int j = 0; j < n>

{

if (a[j] > a[j+1])

{

int temp = a[j];

a[j] = a[j+1];

a[j+1] = temp;

}}}}

int main()

{

int a[20], n, i;

scanf(“%d”, &n);

for(i=0; i

{

scanf(“%d”, &a[i]);

}

sort(a, n);

printf(“The minimum sum of absolute is %d”,abs_sum(a, n));

return 0;

}

C++

// C++ program to find the sum of minimum absolute difference of the given array

#include

using namespace std;

int abs_sum(int a[], int len);

int main()

{

int a[20], n, i;

cout << “\nEnter the number of elements : “;

cin >> n;

cout << “\nInput the array elements : “;

for(i=0; i

{

cin >> a[i];

}

cout << “\nThe sum of minimum absolute differences : ” << abs>

return 0;

}

// Function to calculate absolute sum

int abs_sum(int arr[], int n)

{

sort(arr, arr+n);

int sum = 0;

sum += abs(arr[0] – arr[1]);

sum += abs(arr[n-1] – arr[n-2]);

for (int i=1; i

sum += min(abs(arr[i] – arr[i-1]), abs(arr[i] – arr[i+1]));  // Total sum of absolute differences

return sum;

}

JAVA

// Java program to find the sum of minimum absolute difference in an array

import java.*;

import java.util.Arrays;

public class Main

{

static int sumOfMinAbsDifferences(

int arr[] ,int n)

{

Arrays.sort(arr);

int sum = 0;

sum += Math.abs(arr[0] – arr[1]);

sum += Math.abs(arr[n-1] – arr[n-2]);

for (int i = 1; i < n>

sum +=

Math.min(Math.abs(arr[i] – arr[i-1]),

Math.abs(arr[i] – arr[i+1])); // Total sum of minimum absolute difference

return sum;

}

public static void main(String args[])

{

int arr[] = {1, 3, 6, 9, 3};

int n = arr.length;

System.out.println( “Sum = ” + sumOfMinAbsDifferences(arr, n));

}

}

PYTHON 3

# Python program to find the sum of minimum absolute differences in an array

def sumOfMinAbsDifferences(arr,n):

arr.sort()

sum = 0

sum += abs(arr[0] – arr[1]);

sum += abs(arr[n – 1] – arr[n – 2]);

for i in range(1, n – 1):

sum += min(abs(arr[i] – arr[i – 1]),

abs(arr[i] – arr[i + 1]))  // Total sum of minimum absolute difference

return sum;

arr = [1,2,3,4,5]

n = len(arr)

print( “Sum = “, sumOfMinAbsDifferences(arr, n))

Output

Enter the number of elements : 5

Input the array elements : 1 3 9 6 3

The minimum sum of absolute is : 11

Time complexity: O(n)