Product of absolute difference of every pair in given Array

Given an array arr[] of N elements, the task is to find the product of absolute differences of all pairs in the given array.

Examples:

Input: arr[] = {1, 2, 3, 4} 
Output: 12
Explanation:
Product of |2-1| * |3-1| * |4-1| * |3-2| * |4-2| * |4-3| = 12
Input: arr[] = {1, 8, 9, 15, 16} 
Output: 27659520 

Approach: The idea is to generate every possible pairs of the given array arr[] and find the product of the absolute difference of all the pairs. 

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the product of
// abs diff of all pairs (x, y)
int getProduct(int a[], int n)
{
    // To store product
    int p = 1;

    // Iterate all possible pairs
    for (int i = 0; i < n; i++) {

        for (int j = i + 1; j < n; j++) {

            // Find the product
            p *= abs(a[i] - a[j]);
        }
    }

    // Return product
    return p;
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << getProduct(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to return the product of
// abs diff of all pairs (x, y)
static int getProduct(int a[], int n)
{

    // To store product
    int p = 1;

    // Iterate all possible pairs
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {

            // Find the product
            p *= Math.abs(a[i] - a[j]);
        }
    }

    // Return product
    return p;
}

// Driver Code
public static void main(String[] args)
{

    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int N = arr.length;

    // Function call
    System.out.println(getProduct(arr, N));
}
}

// This code is contributed by Ritik Bansal

Python3

# Python3 program for
# the above approach

# Function to return the product of
# abs diff of all pairs (x, y)
def getProduct(a, n):

    # To store product
    p = 1

    # Iterate all possible pairs
    for i in range (n):
        for j in range (i + 1, n):

            # Find the product
            p *= abs(a[i] - a[j])

    # Return product
    return p

# Driver Code
if __name__ == "__main__":

    # Given array arr[]
    arr = [1, 2, 3, 4]
    N = len(arr)

    # Function Call
    print (getProduct(arr, N))

# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Function to return the product of
// abs diff of all pairs (x, y)
static int getProduct(int []a, int n)
{

    // To store product
    int p = 1;

    // Iterate all possible pairs
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {

            // Find the product
            p *= Math.Abs(a[i] - a[j]);
        }
    }

    // Return product
    return p;
}

// Driver Code
public static void Main(string[] args)
{

    // Given array arr[]
    int []arr = { 1, 2, 3, 4 };
    int N = arr.Length;

    // Function call
    Console.Write(getProduct(arr, N));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// Javascript program for the above approach

// Function to return the product of
// abs diff of all pairs (x, y)
function getProduct( a, n)
{
    // To store product
    var p = 1;

    // Iterate all possible pairs
    for (var i = 0; i < n; i++) {

        for (var j = i + 1; j < n; j++) {

            // Find the product
            p *= Math.abs(a[i] - a[j]);
        }
    }

    // Return product
    return p;
}

// Driver Code

// Given array arr[]
var arr = [ 1, 2, 3, 4 ];
var N = arr.length;

// Function Call
document.write( getProduct(arr, N));

// This code is contributed by itsok.
</script>

Output:

12

Time Complexity: O(N²) 
Auxiliary Space: O(1)