# 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 ` `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

 ``

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 ` `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

 ``

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

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