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: {3, 4, 5} = 3 + 4 + 5 = 12 
Subarray 4: {4, 5, 6} = 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++ implementation 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;
}

Java

// Java implementation 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

C#

// C# implementation 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

Python3

# Python3 implementation 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]

        # Print sum 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

Javascript

<script>

// JavaScript implementation to find the sum
// of all subarrays of size K

// Function to find the sum of
// all subarrays of size K
function calcSum(arr, n, k)
{

    // Loop to consider every
    // subarray of size K
    for (var i = 0; i <= n - k; i++) {

        // Initialize sum = 0
        var sum = 0;

        // Calculate sum of all elements
        // of current subarray
        for (var j = i; j < k + i; j++)
            sum += arr[j];

        // Print sum of each subarray
        document.write(sum + " ");
    }
}

// Driver Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var n = arr.length;
var k = 3;

// Function Call
calcSum(arr, n, k);


</script>

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++ implementation 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 subtract
        // 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;
}

Java

// Java implementation 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 subtract
        // 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

Python3

# Python3 implementation 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 subtract
        # 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

C#

// C# implementation 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 subtract
        // 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

Javascript

<script>

// Javascript implementation to find the sum
// of all subarrays of size K

// Function to find the sum of
// all subarrays of size K
function calcSum(arr, n, k)
{

    // Initialize sum = 0
    var sum = 0;

    // Consider first subarray of size k
    // Store the sum of elements
    for (var i = 0; i < k; i++)
        sum += arr[i];

    // Print the current sum
    document.write( sum + " ");

    // Consider every subarray of size k
    // Remove first element and add current
    // element to the window
    for (var i = k; i < n; i++) {

        // Add the element which enters
        // into the window and subtract
        // the element which pops out from
        // the window of the size K
        sum = (sum - arr[i - k]) + arr[i];

        // Print the sum of subarray
        document.write( sum + " ");
    }
}

// Drivers Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var n = arr.length;
var k = 3;

// Function Call
calcSum(arr, n, k);

// This code is contributed by noob2000.
</script>

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).
  

Related Topic: Subarrays, Subsequences, and Subsets in Array