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Sum of absolute differences of pairs from the given array that satisfy the given condition
  • Last Updated : 13 Sep, 2019

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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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
Output:
45

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