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

  • Last Updated : 18 Nov, 2021

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

Javascript




<script>
 
// JS implementation of the approach
 
// Function that returns true
// if n is prime
function isPrime(n)
{
 
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for (let 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
function findSum( arr, n)
{
 
    // To store the required sum
    let sum = 0;
 
    for (let i = 0; i < n - 1; i++) {
        for (let 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 + Math.abs(arr[i] - arr[j]);
            }
    }
 
    // Return the sum
    return sum;
}
 
// Driver code
let arr = [ 1, 2, 3, 5, 7, 12 ];
let n = arr.length;
document.write(findSum(arr, n));
</script>
Output
45

Time Complexity: O(N3)

Auxiliary Space: O(1)


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