# 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: 10
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 ` `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); `   `    ``// 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`

Output:

```12

```

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

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