Sum of all subarrays of size K

Given an array arr[] and an integer K, the task is to calculate the sum of all subarrays of size K.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 3
Output: 6 9 12 15
Explanation:
All subarrays of size k and their sum:
Subarray 1: {1, 2, 3} = 1 + 2 + 3 = 6
Subarray 2: {2, 3, 4} = 2 + 3 + 4 = 9
Subarray 3: {1, 2, 3} = 3 + 4 + 5 = 12
Subarray 4: {1, 2, 3} = 4 + 5 + 6 = 15

Input: arr[] = {1, -2, 3, -4, 5, 6}, K = 2
Output: -1, 1, -1, 1, 11
Explanation:
All subarrays of size K and their sum:
Subarray 1: {1, -2} = 1 – 2 = -1
Subarray 2: {-2, 3} = -2 + 3 = -1
Subarray 3: {3, 4} = 3 – 4 = -1
Subarray 4: {-4, 5} = -4 + 5 = 1
Subarray 5: {5, 6} = 5 + 6 = 11

Naive Approach: The naive approach will be to generate all subarrays of size K and find the sum of each subarray using iteration.



Below is the implementation of the above approach:

C/C++

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// C++ implementaion to find the sum
// of all subarrays of size K
  
#include <iostream>
using namespace std;
  
// Function to find the sum of 
// all subarrays of size K
int calcSum(int arr[], int n, int k)
{
  
    // Loop to consider every 
    // subarray of size K
    for (int i = 0; i <= n - k; i++) {
          
        // Initialize sum = 0
        int sum = 0;
  
        // Calculate sum of all elements
        // of current subarray
        for (int j = i; j < k + i; j++)
            sum += arr[j];
  
        // Print sum of each subarray
        cout << sum << " ";
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
  
    // Function Call
    calcSum(arr, n, k);
  
    return 0;
}

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Java

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// Java implementaion to find the sum
// of all subarrays of size K
class GFG{
   
// Function to find the sum of 
// all subarrays of size K
static void calcSum(int arr[], int n, int k)
{
   
    // Loop to consider every 
    // subarray of size K
    for (int i = 0; i <= n - k; i++) {
           
        // Initialize sum = 0
        int sum = 0;
   
        // Calculate sum of all elements
        // of current subarray
        for (int j = i; j < k + i; j++)
            sum += arr[j];
   
        // Print sum of each subarray
        System.out.print(sum+ " ");
    }
}
   
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;
    int k = 3;
   
    // Function Call
    calcSum(arr, n, k); 
}
}
  
// This code is contributed by Rajput-Ji

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C#

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// C# implementaion to find the sum
// of all subarrays of size K
using System; 
  
class GFG  
    
    // Function to find the sum of 
    // all subarrays of size K
    static  void calcSum(int[] arr, int n, int k)
    {
      
        // Loop to consider every 
        // subarray of size K
        for (int i = 0; i <= n - k; i++) {
              
            // Initialize sum = 0
            int sum = 0;
      
            // Calculate sum of all elements
            // of current subarray
            for (int j = i; j < k + i; j++)
                sum += arr[j];
      
            // Print sum of each subarray
            Console.Write(sum + " ");
        }
    }
      
    // Driver Code
    static void Main() 
    {
        int[] arr = new int[] { 1, 2, 3, 4, 5, 6 };
        int n = arr.Length;
        int k = 3;
      
        // Function Call
        calcSum(arr, n, k);
      
    }
}
  
// This code is contributed by shubhamsingh10

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Python3

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# Python3 implementaion to find the sum
# of all subarrays of size K
   
# Function to find the sum of 
# all subarrays of size K
def calcSum(arr, n, k):
   
    # Loop to consider every 
    # subarray of size K
    for i in range(n - k + 1):
           
        # Initialize sum = 0
        sum = 0
   
        # Calculate sum of all elements
        # of current subarray
        for j in range(i, k + i):
            sum += arr[j]
   
        # Prsum of each subarray
        print(sum, end=" ")
   
# Driver Code
arr=[1, 2, 3, 4, 5, 6]
n = len(arr)
k = 3
  
# Function Call
calcSum(arr, n, k)
  
# This code is contributed by mohit kumar 29

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Output:

6 9 12 15

Performance Analysis:

  • Time Complexity: As in the above approach, There are two loops, where first loop runs (N – K) times and second loop runs for K times. Hence the Time Complexity will be O(N*K).
  • Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).

Efficient Approach: Using Sliding Window The idea is to use the sliding window approach to find the sum of all possible subarrays in the array.

  • For each size in the range [0, K], find the sum of the first window of size K and store it in an array.
  • Then for each size in the range [K, N], add the next element which contributes into the sliding window and subtract the element which pops out from the window.
    // Adding the element which
    // adds into the new window
    sum = sum + arr[j]
    
    // Subtracting the element which
    // pops out from the window
    sum = sum - arr[j-k]
    
    where sum is the variable to store the result
          arr is the given array
          j is the loop variable in range [K, N]
    

Below is the implementation of the above approach:

C/C++

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// C++ implementaion to find the sum
// of all subarrays of size K
  
#include <iostream>
using namespace std;
  
// Function to find the sum of 
// all subarrays of size K
int calcSum(int arr[], int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
  
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
  
    // Print the current sum
    cout << sum << " ";
  
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
          
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
          
        // Print the sum of subarray
        cout << sum << " ";
    }
}
  
// Drivers Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
      
    // Function Call
    calcSum(arr, n, k);
  
    return 0;
}

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Java

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// Java implementaion to find the sum
// of all subarrays of size K
class GFG{
  
// Function to find the sum of 
// all subarrays of size K
static void calcSum(int arr[], int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
  
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
  
    // Print the current sum
    System.out.print(sum+ " ");
  
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
          
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
          
        // Print the sum of subarray
        System.out.print(sum+ " ");
    }
}
  
// Drivers Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;
    int k = 3;
      
    // Function Call
    calcSum(arr, n, k);
}
}
  
// This code is contributed by sapnasingh4991

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Python3

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# Python3 implementaion to find the sum
# of all subarrays of size K
   
# Function to find the sum of 
# all subarrays of size K
def calcSum(arr,  n, k):
  
    # Initialize sum = 0
    sum = 0
   
    # Consider first subarray of size k
    # Store the sum of elements
    for i in range( k):
        sum += arr[i]
   
    # Print the current sum
    print( sum ,end= " ")
   
    # Consider every subarray of size k
    # Remove first element and add current
    # element to the window
    for i in range(k,n):
           
        # Add the element which enters
        # into the window and substract
        # the element which pops out from
        # the window of the size K
        sum = (sum - arr[i - k]) + arr[i]
           
        # Print the sum of subarray
        print( sum ,end=" ")
   
# Drivers Code
if __name__ == "__main__":
  
    arr = [ 1, 2, 3, 4, 5, 6 ]
    n = len(arr)
    k = 3
       
    # Function Call
    calcSum(arr, n, k)
   
# This code is contributed by chitranayal

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C#

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// C# implementaion to find the sum
// of all subarrays of size K
using System;
  
class GFG{
   
// Function to find the sum of 
// all subarrays of size K
static void calcSum(int []arr, int n, int k)
{
    // Initialize sum = 0
    int sum = 0;
   
    // Consider first subarray of size k
    // Store the sum of elements
    for (int i = 0; i < k; i++)
        sum += arr[i];
   
    // Print the current sum
    Console.Write(sum+ " ");
   
    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (int i = k; i < n; i++) {
           
        // Add the element which enters
        // into the window and substract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];
           
        // Print the sum of subarray
        Console.Write(sum + " ");
    }
}
   
// Drivers Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    int n = arr.Length;
    int k = 3;
       
    // Function Call
    calcSum(arr, n, k);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

6 9 12 15

Performance Analysis:

  • Time Complexity: As in the above approach, There is one loop which take O(N) time. Hence the Time Complexity will be O(N).
  • Auxiliary Space Complexity: As in the above approach, There is no extra space used. Hence the auxiliary space complexity will be O(1).

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