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.

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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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