# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` ` `  `    ``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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