Sum of absolute differences of pairs from the given array that satisfy the given condition

Given an array arr[] of N elements, the task is to find the sum of absolute differences between all pairs (arr[i], arr[j]) such that i < j and (j – i) is prime.

Example:

Input: arr[] = {1, 2, 3, 5, 7, 12}
Output: 45
All valid index pairs are:
(5, 0) -> abs(12 – 1) = 11
(3, 0) -> abs(5 – 1) = 4
(2, 0) -> abs(3 – 1) = 2
(4, 1) -> abs(7 – 2) = 5
(3, 1) -> abs(5 – 2) = 3
(5, 2) -> abs(12 – 3) = 9
(4, 2) -> abs(7 – 3) = 4
(5, 3) -> abs(12 – 5) = 7
11 + 4 + 2 + 5 + 3 + 9 + 4 + 7 = 45



Input: arr[] = {2, 5, 6, 7}
Output: 11

Approach: Initialise sum = 0 and run two nested loops and for every pair arr[i], arr[j] is (j – i) is prime then update the sum as sum = sum + abs(arr[i], arr[j]). Print the sum in the end.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function that returns true
// if n is prime
bool isPrime(int n)
{
  
    // Corner case
    if (n <= 1)
        return false;
  
    // Check from 2 to n-1
    for (int i = 2; i < n; i++)
        if (n % i == 0)
            return false;
  
    return true;
}
  
// Function to return the absolute
// differences of the pairs which
// satisfy the given condition
int findSum(int arr[], int n)
{
  
    // To store the required sum
    int sum = 0;
  
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++)
  
            // If difference between the indices
            // is prime
            if (isPrime(j - i)) {
  
                // Update the sum with the absolute
                // difference of the pair elements
                sum = sum + abs(arr[i] - arr[j]);
            }
    }
  
    // Return the sum
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 5, 7, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << findSum(arr, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
    // Function that returns true
    // if n is prime
    static boolean isPrime(int n) 
    {
  
        // Corner case
        if (n <= 1)
        {
            return false;
        }
  
        // Check from 2 to n-1
        for (int i = 2; i < n; i++) 
        {
            if (n % i == 0
            {
                return false;
            }
        }
        return true;
    }
  
    // Function to return the absolute
    // differences of the pairs which
    // satisfy the given condition
    static int findSum(int arr[], int n) 
    {
  
        // To store the required sum
        int sum = 0;
  
        for (int i = 0; i < n - 1; i++) 
        {
            // If difference between the indices is prime
            for (int j = i + 1; j < n; j++) 
            {
                if (isPrime(j - i)) 
                {
  
                    // Update the sum with the absolute
                    // difference of the pair elements
                    sum = sum + Math.abs(arr[i] - arr[j]);
                }
            }
        }
  
        // Return the sum
        return sum;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int arr[] = {1, 2, 3, 5, 7, 12};
        int n = arr.length;
  
        System.out.println(findSum(arr, n));
    }
  
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function that returns true 
# if n is prime 
def isPrime(n) : 
  
    # Corner case 
    if (n <= 1) :
        return False
  
    # Check from 2 to n-1 
    for i in range(2, n) :
        if (n % i == 0) :
            return False
  
    return True
  
# Function to return the absolute 
# differences of the pairs which 
# satisfy the given condition 
def findSum(arr, n) : 
  
    # To store the required sum 
    sum = 0
  
    for i in range(n - 1) :
        for j in range(i + 1, n) : 
  
            # If difference between the indices 
            # is prime 
            if (isPrime(j - i)) :
  
                # Update the sum with the absolute 
                # difference of the pair elements 
                sum = sum + abs(arr[i] - arr[j]); 
  
    # Return the sum 
    return sum
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 2, 3, 5, 7, 12 ];
    n = len(arr); 
  
    print(findSum(arr, n)); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG 
{
  
    // Function that returns true
    // if n is prime
    static bool isPrime(int n) 
    {
  
        // Corner case
        if (n <= 1)
        {
            return false;
        }
  
        // Check from 2 to n-1
        for (int i = 2; i < n; i++) 
        {
            if (n % i == 0) 
            {
                return false;
            }
        }
        return true;
    }
  
    // Function to return the absolute
    // differences of the pairs which
    // satisfy the given condition
    static int findSum(int []arr, int n) 
    {
  
        // To store the required sum
        int sum = 0;
  
        for (int i = 0; i < n - 1; i++) 
        {
            // If difference between the indices is prime
            for (int j = i + 1; j < n; j++) 
            {
                if (isPrime(j - i)) 
                {
  
                    // Update the sum with the absolute
                    // difference of the pair elements
                    sum = sum + Math.Abs(arr[i] - arr[j]);
                }
            }
        }
  
        // Return the sum
        return sum;
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int []arr = {1, 2, 3, 5, 7, 12};
        int n = arr.Length;
  
        Console.WriteLine(findSum(arr, n));
    }
  
// This code is contributed by PrinciRaj1992

chevron_right


Output:

45


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.