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++ 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 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 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# 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 |
<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> |
45
Time Complexity: O(N3)
Auxiliary Space: O(1)