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# Sum of all subarrays of size K

• Difficulty Level : Easy
• Last Updated : 17 Aug, 2021

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);``    ``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]` `        ``# 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`

## 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);``    ``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).

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