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: 

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).

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: 

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