# Product of all Subarrays of an Array | Set 2

Given an array arr[] of integers of size N, the task is to find the products of all subarrays of the array.

Examples:

Input: arr[] = {2, 4}
Output: 64
Explanation:
Here, subarrays are {2}, {2, 4}, and {4}.
Products of each subarray are 2, 8, 4.
Product of all Subarrays = 64

Input: arr[] = {1, 2, 3}
Output: 432
Explanation:
Here, subarrays are {1}, {1, 2}, {1, 2, 3}, {2}, {2, 3}, {3}.
Products of each subarray are 1, 2, 6, 2, 6, 3.
Product of all Subarrays = 432

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive and Iterative approach: Please refer this post for these approaches.

Approach: The idea is to count the number of each element occurs in all the subarrays. To count we have below observations:

• In every subarray beginning with arr[i], there are (N – i) such subsets starting with the element arr[i].
For Example:

For array arr[] = {1, 2, 3}
N = 3 and for element 2 i.e., index = 1
There are (N – index) = 3 – 1 = 2 subsets
{2} and {2, 3}

• For any element arr[i], there are (N – i)*i subarrays where arr[i] is not the first element.

For array arr[] = {1, 2, 3}
N = 3 and for element 2 i.e., index = 1
There are (N – index)*index = (3 – 1)*1 = 2 subsets where 2 is not the first element.
{1, 2} and {1, 2, 3}

Therefore, from the above observations, the total number of each element arr[i] occurs in all the subarrays at every index i is given by:

```total_elements = (N - i) + (N - i)*i
total_elements = (N - i)*(i + 1)
```

The idea is to multiply each element (N – i)*(i + 1) number of times to get the product of elements in all subarrays.

Below is the implementation of the above approach:

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the product of ` `// elements of all subarray ` `long` `int` `SubArrayProdct(``int` `arr[], ` `                        ``int` `n) ` `{ ` `    ``// Initialize the result ` `    ``long` `int` `result = 1; ` ` `  `    ``// Computing the product of ` `    ``// subarray using formula ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``result *= ``pow``(arr[i], ` `                      ``(i + 1) * (n - i)); ` ` `  `    ``// Return the product of all ` `    ``// elements of each subarray ` `    ``return` `result; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array arr[] ` `    ``int` `arr[] = { 2, 4 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``cout << SubArrayProdct(arr, N) ` `         ``<< endl; ` `    ``return` `0; ` `} `

 `// Java program for the above approach  ` `import` `java.util.*;  ` ` `  `class` `GFG{  ` ` `  `// Function to find the product of ` `// elements of all subarray ` `static` `int` `SubArrayProdct(``int` `arr[], ``int` `n) ` `{ ` `     `  `    ``// Initialize the result ` `    ``int` `result = ``1``; ` ` `  `    ``// Computing the product of ` `    ``// subarray using formula ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `       ``result *= Math.pow(arr[i], (i + ``1``) * ` `                                  ``(n - i)); ` ` `  `    ``// Return the product of all ` `    ``// elements of each subarray ` `    ``return` `result; ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` ` `  `    ``// Given array arr[] ` `    ``int` `arr[] = ``new` `int``[]{``2``, ``4``}; ` ` `  `    ``int` `N = arr.length; ` ` `  `    ``// Function Call ` `    ``System.out.println(SubArrayProdct(arr, N));  ` `}  ` `}  ` ` `  `// This code is contributed by Pratima Pandey  `

 `# Python3 program for the above approach ` ` `  `# Function to find the product of ` `# elements of all subarray ` `def` `SubArrayProdct(arr, n): ` ` `  `    ``# Initialize the result ` `    ``result ``=` `1``; ` ` `  `    ``# Computing the product of ` `    ``# subarray using formula ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``result ``*``=` `pow``(arr[i], ` `                     ``(i ``+` `1``) ``*` `(n ``-` `i)); ` ` `  `    ``# Return the product of all ` `    ``# elements of each subarray ` `    ``return` `result; ` ` `  `# Driver Code ` ` `  `# Given array arr[] ` `arr ``=` `[ ``2``, ``4` `]; ` `N ``=` `len``(arr); ` ` `  `# Function Call ` `print``(SubArrayProdct(arr, N)) ` ` `  `# This code is contributed by Code_Mech `

 `// C# program for the above approach  ` `using` `System; ` `class` `GFG{  ` ` `  `// Function to find the product of ` `// elements of all subarray ` `static` `int` `SubArrayProdct(``int` `[]arr, ``int` `n) ` `{ ` `     `  `    ``// Initialize the result ` `    ``int` `result = 1; ` ` `  `    ``// Computing the product of ` `    ``// subarray using formula ` `    ``for``(``int` `i = 0; i < n; i++) ` `       ``result *= (``int``)(Math.Pow(arr[i], (i + 1) * ` `                                        ``(n - i))); ` ` `  `    ``// Return the product of all ` `    ``// elements of each subarray ` `    ``return` `result; ` `} ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` ` `  `    ``// Given array arr[] ` `    ``int` `[]arr = ``new` `int``[]{2, 4}; ` ` `  `    ``int` `N = arr.Length; ` ` `  `    ``// Function Call ` `    ``Console.Write(SubArrayProdct(arr, N));  ` `}  ` `}  ` ` `  `// This code is contributed by Code_Mech `

Output:
```64
```

Time Complexity: O(N), where N is the number of elements.
Auxiliary Space: O(1)

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Improved By : dewantipandeydp, Code_Mech

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