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#

   // 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 non repeating Subarrays of an Array

spp____

Improve

Article Tags :

Practice Tags :

Product of all Subarrays of an Array | Set 2

C++

Java

Python3

C#

Javascript

Similar Reads