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 = 15Input: 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++ 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 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# 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 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 |
<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> |
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++ 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 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 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# 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 |
<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> |
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