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



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

C/C++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

210

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++

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// 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;
}

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

210

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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Improved By : Sam007, 29AjayKumar, nidhi_biet



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