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:64Explanation:

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:1944810000Explanation:

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` `occurance = ` `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 occurance` ` ` `// and then taking product of those.` ` ` `product *= ` `pow` `(arr[i], occurance);` ` ` `}` ` ` `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` `occurance = 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 occurance` ` ` `// and then taking product of those.` ` ` `product *= Math.pow(arr[i], occurance);` ` ` `}` ` ` ` ` `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` ` ` `occurance ` `=` `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 occurance` ` ` `# and then taking product of those.` ` ` `product ` `*` `=` `pow` `(arr[i], occurance);` ` ` ` ` `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` `occurance = 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 occurance` ` ` `// and then taking product of those.` ` ` `product *= Math.Pow(arr[i], occurance);` ` ` `}` ` ` `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` |

**Output:**

1944810000

**Time Complexity:** *O(N)*, where N is the size of the array.

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