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Product of all Subarrays of an Array | Set 2
  • Difficulty Level : Medium
  • 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: 
64

 

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

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