Product of all non repeating Subarrays of an Array

Given an array containing distinct integers arr[] of size N, the task is to print the product of all non-repeating subarrays of the array. 
Examples:

Input: arr[] = {2, 4} 
Output: 64 
Explanation:
The possible subarrays for the given array are {2}, {2, 4}, {4} 
The products are 2, 8, 4 respectively. Therefore, the overall product of all the subarrays = 64
Input: arr[] = {10, 3, 7} 
Output: 1944810000 
Explanation:
The possible subarrays for the given array are {10}, {10, 3}, {0, 7}, {10, 3, 7}, {3}, {7}, {3, 7}. 
The products are 10, 30, 70, 210, 3, 7, 21 respectively. Therefore, the overall product of all the subarrays = 1944810000 

Naive Approach: The naive approach for this problem is to generate all the sub-arrays of the given array and compute their product. The time complexity of this approach is exponential. 
Efficient Approach: The idea is to make an observation. If we observe the non-repeating sub-arrays, we can observe that the occurrences of a single element in the sub-arrays follows a relationship with the length of the array. 

• For example, let the array arr[] = {10, 3, 7}.
• All the non repeating possible subarrays of the above array are: {{10}, {10, 3}, {10, 7}, {10, 3, 7}, {3}, {7}, {3, 7}}.
• In the above sub-arrays, the frequency of every element can be observed as: 
  

Frequency of 10 in subarrays = 4
Frequency of 3 in subarrays = 4
Frequency of 7 in subarrays = 4

• Here, the following identity holds true for the arrays of any length: 
  

Frequency of element = 2(arr.length-1)

• Hence, in order to get the final product, we can simply perform: 
  

product = 10 * (freq_of_10) * 3 * (freq_of_3) * 7 * (freq_of_7)

• Therefore, the idea is to simply iterate through the array and perform multiplication of every element and its corresponding frequency.

Below is the implementation of the above approach: 

C++

// C++ program to find the product of
// all non-repeating Subarrays of an Array
#include <bits/stdc++.h>
using namespace std;

// Function to find the product of
// all non-repeating Subarrays of an Array
long product(int arr[], int n)
{

    // Finding the occurrence of every element
    double occurrence = pow(2, n - 1);

    double product = 1;

    // Iterating through the array and
    // finding the product
    for(int i = 0; i < n; i++)
    {

        // We are taking the power of each
        // element in array with the occurrence
        // and then taking product of those.
        product *= pow(arr[i], occurrence);
    }
    return (long)product;
}

// Driver code
int main()
{
    int arr[] = { 10, 3, 7 };

    int len = sizeof(arr) / sizeof(arr[0]);

    cout << product(arr, len);
    return 0;
}

// This code is contributed by PrinciRaj1992

Java

// Java program to find the product of
// all non-repeating Subarrays of an Array

public class GFG {

    // Function to find the product of
    // all non-repeating Subarrays of an Array
    private static long product(int[] arr)
    {
        // Finding the occurrence of every element
        double occurrence = Math.pow(2, arr.length - 1);

        double product = 1;

        // Iterating through the array and
        // finding the product
        for (int i = 0; i < arr.length; i++) {

            // We are taking the power of each
            // element in array with the occurrence
            // and then taking product of those.
            product *= Math.pow(arr[i], occurrence);
        }

        return (long)product;
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 10, 3, 7 };

        System.out.println(product(arr));
    }
}

Python3

# Python3 program to find the product of
# all non-repeating Subarrays of an Array

# Function to find the product of
# all non-repeating Subarrays of an Array
def product(arr):

    # Finding the occurrence of every element
    occurrence = pow(2, len(arr) - 1);

    product = 1;

    # Iterating through the array and
    # finding the product
    for i in range(0, len(arr)):

        # We are taking the power of each
        # element in array with the occurrence
        # and then taking product of those.
        product *= pow(arr[i], occurrence);

    return product;

# Driver code
arr = [ 10, 3, 7 ];
print(product(arr));

# This code is contributed by Code_Mech

C#

// C# program to find the product of
// all non-repeating Subarrays of an Array
using System;
class GFG{

// Function to find the product of
// all non-repeating Subarrays of an Array
private static long product(int[] arr)
{
    // Finding the occurrence of every element
    double occurrence = Math.Pow(2, arr.Length - 1);

    double product = 1;

    // Iterating through the array and
    // finding the product
    for (int i = 0; i < arr.Length; i++)
    {

        // We are taking the power of each
        // element in array with the occurrence
        // and then taking product of those.
        product *= Math.Pow(arr[i], occurrence);
    }
    return (long)product;
}

// Driver code
public static void Main(String[] args)
{
    int[] arr = { 10, 3, 7 };

    Console.WriteLine(product(arr));
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// Javascript program to find the product of
// all non-repeating Subarrays of an Array

    // Function to find the product of
    // all non-repeating Subarrays of an Array
    function product(arr)
    {
        // Finding the occurrence of every element
        let occurrence = Math.pow(2, arr.length - 1);

        let product = 1;

        // Iterating through the array and
        // finding the product
        for (let i = 0; i < arr.length; i++) {

            // We are taking the power of each
            // element in array with the occurrence
            // and then taking product of those.
            product *= Math.pow(arr[i], occurrence);
        }

        return product;
    }


// Driver Code

    let arr = [ 10, 3, 7 ];

    document.write(product(arr));

</script>

Output:

1944810000

Time Complexity: O(N), where N is the size of the array 
Auxiliary Space: O(1)