Product of all Subarrays of an Array | Set 2

Last Updated : 05 May, 2021

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

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++

// C++ program for the above approach #include <bits/stdc++.h> 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[0]);      // Function Call     cout << SubArrayProdct(arr, N)          << endl;     return 0; }

Java

// 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

# 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

JavaScript

<script>  // JavaScript program to implement // the above approach  // Function to find the product of // elements of all subarray function SubArrayProdct(arr, n) {            // Initialize the result     let result = 1;        // Computing the product of     // subarray using formula     for(let 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       // Given array arr[]     let arr = [2, 4];        let N = arr.length;        // Function Call     document.write(SubArrayProdct(arr, N));     // This code is contributed by sanjoy_62. </script>

Output:

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

Product of all Subarrays of an Array | Set 2

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Product of all Subarrays of an Array | Set 2

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