Sum of product of all elements of sub-arrays of size k

Given an array and a number k, the task is to calculate the sum of the product of all elements of subarrays of size k.

Examples :

Input : arr[] = {1, 2, 3, 4, 5, 6} 
                k = 3
Output : 210
Consider all subarrays of size k
1*2*3 = 6
2*3*4 = 24
3*4*5 = 60
4*5*6 = 120
6 + 24 + 60 + 120 = 210

Input : arr[] = {1, -2, 3, -4, 5, 6} 
            k = 2
Output : -10
Consider all subarrays of size k
1*-2 = -2
-2*3 = -6
3*-4 = -12
-4*5 = -20
5*6  =   30
-2 + -6 + -12 + -20+ 30 = -10

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

A Naive approach is to generate all subarrays of size k and do the sum of product of all elements of subarrays.

C/C++

// C++ program to find the sum of 
// product of all subarrays 
#include <iostream> 
using namespace std; 
  
// Function to calculate the sum of product 
int calcSOP(int arr[], int n, int k) 
{ 
    // Initialize sum = 0 
    int sum = 0; 
  
    // Consider every subarray of size k 
    for (int i = 0; i <= n - k; i++) { 
        int prod = 1; 
  
        // Calculate product of all elements 
        // of current subarray 
        for (int j = i; j < k + i; j++) 
            prod *= arr[j]; 
  
        // Store sum of all the products 
        sum += prod; 
    } 
  
    // Return sum 
    return sum; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 1, 2, 3, 4, 5, 6 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int k = 3; 
  
    cout << calcSOP(arr, n, k); 
  
    return 0; 
} 

Java

// Java program to find the sum of 
// product of all subarrays 
  
class GFG { 
    // Method to calculate the sum of product 
    static int calcSOP(int arr[], int n, int k) 
    { 
        // Initialize sum = 0 
        int sum = 0; 
  
        // Consider every subarray of size k 
        for (int i = 0; i <= n - k; i++) { 
            int prod = 1; 
  
            // Calculate product of all elements 
            // of current subarray 
            for (int j = i; j < k + i; j++) 
                prod *= arr[j]; 
  
            // Store sum of all the products 
            sum += prod; 
        } 
  
        // Return sum 
        return sum; 
    } 
  
    // Driver method 
    public static void main(String[] args) 
    { 
        int arr[] = { 1, 2, 3, 4, 5, 6 }; 
        int k = 3; 
  
        System.out.println(calcSOP(arr, arr.length, k)); 
    } 
} 

Python3

# python program to find the sum of 
# product of all subarrays 
  
# Function to calculate the sum of product 
def calcSOP(arr, n, k): 
      
    # Initialize sum = 0 
    sum = 0
  
    # Consider every subarray of size k 
    for i in range(0, (n-k)+1): 
        prod = 1
          
        # Calculate product of all elements 
        # of current subarray 
        for j in range(i, k+i): 
            prod = int(prod * arr[j]) 
  
        # Store sum of all the products 
        sum = sum + prod 
      
    # Return sum 
    return sum
  
#Driver code 
arr = [ 1, 2, 3, 4, 5, 6 ] 
n = len(arr) 
k = 3
print(calcSOP(arr, n, k)) 
  
# This code is contributed by Sam007 

C#

// C# program to find the sum of 
// product of all subarrays 
using System; 
  
public class GFG { 
      
    // Method to calculate the sum of product 
    static int calcSOP(int[] arr, int n, int k) 
    { 
        // Initialize sum = 0 
        int sum = 0; 
  
        // Consider every subarray of size k 
        for (int i = 0; i <= n - k; i++) { 
            int prod = 1; 
  
            // Calculate product of all elements 
            // of current subarray 
            for (int j = i; j < k + i; j++) 
                prod *= arr[j]; 
  
            // Store sum of all the products 
            sum += prod; 
        } 
  
        // Return sum 
        return sum; 
    } 
  
    // Driver method 
    public static void Main() 
    { 
        int[] arr = {1, 2, 3, 4, 5, 6}; 
        int k = 3; 
  
        Console.WriteLine(calcSOP(arr, arr.Length, k)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP program to find the sum of 
// product of all subarrays 
  
// Function to calculate 
// the sum of product 
function calcSOP($arr, $n, $k) 
{ 
      
    // Initialize sum = 0 
    $sum = 0; 
  
    // Consider every subarray  
    // of size k 
    for ($i = 0; $i <= $n - $k; $i++) 
    { 
        $prod = 1; 
  
        // Calculate product of all  
        // elements of current subarray 
        for ($j = $i; $j < $k + $i; $j++) 
            $prod *= $arr[$j]; 
  
        // Store sum of all  
        // the products 
        $sum += $prod; 
    } 
  
    // Return sum 
    return $sum; 
} 
  
    // Driver code 
    $arr = array( 1, 2, 3, 4, 5, 6 ); 
    $n = count($arr); 
    $k = 3; 
  
    echo calcSOP($arr, $n, $k); 
          
// This code is contributed by Sam007 
?> 

Output:

Time Complexity: O(nk)

An Efficient Method is to use the concept of Sliding Window.

1- Consider first subarray/window of size k, do the product of elements and add to the total_sum.

 for (i=0; i < k; i++)
       prod = prod * arr[i];

2- Now, By Using sliding window concept, remove first element of window from the product and add next element to the window. i.e.

for (i =k ; i < n; i++)
 {
     // Removing first element from product  
     prod = prod / arr[i-k]; 

     //  Adding current element to the product
     prod = prod * arr[i];  
     sum += prod;
 }

3- Return sum

C/C++

// C++ program to find the sum of 
// product of all subarrays 
#include <iostream> 
using namespace std; 
  
// Function to calculate the sum of product 
int calcSOP(int arr[], int n, int k) 
{ 
    // Initialize sum = 0 and prod = 1 
    int sum = 0, prod = 1; 
  
    // Consider first subarray of size k 
    // Store the products of elements 
    for (int i = 0; i < k; i++) 
        prod *= arr[i]; 
  
    // Add the product to the sum 
    sum += prod; 
  
    // Consider every subarray of size k 
    // Remove first element and add current 
    // element to the window 
    for (int i = k; i < n; i++) { 
  
        // Divide by the first element 
        // of previous subarray/ window 
        // and product with the current element 
        prod = (prod / arr[i - k]) * arr[i]; 
  
        // Add current product to the sum 
        sum += prod; 
    } 
  
    // Return sum 
    return sum; 
} 
  
// Drivers code 
int main() 
{ 
    int arr[] = { 1, 2, 3, 4, 5, 6 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    int k = 3; 
  
    cout << calcSOP(arr, n, k); 
  
    return 0; 
} 

Java

// Java program to find the sum of 
// product of all subarrays 
  
class GFG { 
    // Method to calculate the sum of product 
    static int calcSOP(int arr[], int n, int k) 
    { 
        // Initialize sum = 0 and prod = 1 
        int sum = 0, prod = 1; 
  
        // Consider first subarray of size k 
        // Store the products of elements 
        for (int i = 0; i < k; i++) 
            prod *= arr[i]; 
  
        // Add the product to the sum 
        sum += prod; 
  
        // Consider every subarray of size k 
        // Remove first element and add current 
        // element to the window 
        for (int i = k; i < n; i++) { 
  
            // Divide by the first element 
            // of previous subarray/ window 
            // and product with the current element 
            prod = (prod / arr[i - k]) * arr[i]; 
  
            // Add current product to the sum 
            sum += prod; 
        } 
  
        // Return sum 
        return sum; 
    } 
  
    // Driver method 
    public static void main(String[] args) 
    { 
        int arr[] = { 1, 2, 3, 4, 5, 6 }; 
        int k = 3; 
  
        System.out.println(calcSOP(arr, arr.length, k)); 
    } 
} 

Python3

# Python3 program to find the sum of 
# product of all subarrays 
  
# Function to calculate the sum of product 
def calcSOP(arr, n, k): 
      
    # Initialize sum = 0 and prod = 1 
    sum = 0
    prod = 1
  
    # Consider first subarray of size k 
    # Store the products of elements 
    for i in range(k): 
        prod *= arr[i] 
  
    # Add the product to the sum 
    sum += prod 
  
    # Consider every subarray of size k 
    # Remove first element and add current 
    # element to the window 
    for i in range(k, n, 1): 
          
        # Divide by the first element of 
        #  previous subarray/ window and  
        # product with the current element 
        prod = (prod / arr[i - k]) * arr[i] 
  
        # Add current product to the sum 
        sum += prod 
  
    # Return sum 
    return int(sum) 
  
# Drivers code 
arr = [1, 2, 3, 4, 5, 6] 
n = len(arr) 
k = 3
  
print(calcSOP(arr, n, k)) 
  
# This code is contributed 29AjayKumar 

C#

// C# program to find the sum of 
// product of all subarrays 
using System; 
  
public class GFG { 
      
    // Method to calculate the sum of product 
    static int calcSOP(int[] arr, int n, int k) 
    { 
        // Initialize sum = 0 and prod = 1 
        int sum = 0, prod = 1; 
  
        // Consider first subarray of size k 
        // Store the products of elements 
        for (int i = 0; i < k; i++) 
            prod *= arr[i]; 
  
        // Add the product to the sum 
        sum += prod; 
  
        // Consider every subarray of size k 
        // Remove first element and add current 
        // element to the window 
        for (int i = k; i < n; i++) { 
  
            // Divide by the first element 
            // of previous subarray/ window 
            // and product with the current element 
            prod = (prod / arr[i - k]) * arr[i]; 
  
            // Add current product to the sum 
            sum += prod; 
        } 
  
        // Return sum 
        return sum; 
    } 
  
    // Driver method 
    public static void Main() 
    { 
        int[] arr = { 1, 2, 3, 4, 5, 6 }; 
        int k = 3; 
          
        // Function calling 
        Console.WriteLine(calcSOP(arr, arr.Length, k)); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
  
// php program to find the sum of 
// product of all subarrays 
  
// Function to calculate the sum of product 
function calcSOP($arr, $n, $k) 
{ 
    // Initialize sum = 0 and prod = 1 
    $sum = 0; 
    $prod = 1; 
   
    // Consider first subarray of size k 
    // Store the products of elements 
    for ($i = 0; $i < $k; $i++) 
        $prod *= $arr[$i]; 
   
    // Add the product to the sum 
    $sum += $prod; 
   
    // Consider every subarray of size k 
    // Remove first element and add current 
    // element to the window 
    for ($i = $k; $i < $n; $i++) { 
   
        // Divide by the first element 
        // of previous subarray/ window 
        // and product with the current element 
        $prod = ($prod / $arr[$i - $k]) * $arr[$i]; 
   
        // Add current product to the sum 
        $sum += $prod; 
    } 
   
    // Return sum 
    return $sum; 
} 
   
// Drivers code 
    $arr = array( 1, 2, 3, 4, 5, 6 ); 
    $n = count($arr); 
    $k = 3; 
   
    echo calcSOP($arr, $n, $k); 
   
// This code is contributed by Sam007 
?> 

Output:

Time Complexity: O(n)

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

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

C/C++

Java

Python3

C#

PHP

C/C++

Java

Python3

C#

PHP

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