# Find maximum (or minimum) sum of a subarray of size k

Given an array of integers and a number k, find maximum sum of a subarray of size k.

Examples :

```Input  : arr[] = {100, 200, 300, 400}
k = 2
Output : 700

Input  : arr[] = {1, 4, 2, 10, 23, 3, 1, 0, 20}
k = 4
Output : 39
We get maximum sum by adding subarray {4, 2, 10, 23}
of size 4.

Input  : arr[] = {2, 3}
k = 3
Output : Invalid
There is no subarray of size 3 as size of whole
array is 2.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A Simple Solution is to generate all subarrays of size k, compute their sums and finally return maximum of all sums. Time complexity of this solution is O(n*k)

An Efficient Solution is based on the fact that sum of a subarray (or window) of size k can be obtained in O(1) time using sum of previous subarray (or window) of size k. Except first subarray of size k, for other subarrays, we compute sum by removing first element of last window and adding last element of current window.

Below is implementation of above idea.

## C

```// O(n) solution for finding maximum sum of
// a subarray of size k
#include <iostream>
using namespace std;

// Returns maximum sum in a subarray of size k.
int maxSum(int arr[], int n, int k)
{
// k must be greater
if (n < k)
{
cout << "Invalid";
return -1;
}

// Compute sum of first window of size k
int res = 0;
for (int i=0; i<k; i++)
res += arr[i];

// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
int curr_sum = res;
for (int i=k; i<n; i++)
{
curr_sum += arr[i] - arr[i-k];
res = max(res, curr_sum);
}

return res;
}

// Driver code
int main()
{
int arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20};
int k = 4;
int n = sizeof(arr)/sizeof(arr[0]);
cout << maxSum(arr, n, k);
return 0;
}
```

## Java

```// JAVA Code for Find maximum (or minimum)
// sum of a subarray of size k
import java.util.*;

class GFG {

// Returns maximum sum in a subarray of size k.
public static int maxSum(int arr[], int n, int k)
{
// k must be greater
if (n < k)
{
System.out.println("Invalid");
return -1;
}

// Compute sum of first window of size k
int res = 0;
for (int i=0; i<k; i++)
res += arr[i];

// Compute sums of remaining windows by
// removing first element of previous
// window and adding last element of
// current window.
int curr_sum = res;
for (int i=k; i<n; i++)
{
curr_sum += arr[i] - arr[i-k];
res = Math.max(res, curr_sum);
}

return res;
}

/* Driver program to test above function */
public static void main(String[] args)
{
int arr[] = {1, 4, 2, 10, 2, 3, 1, 0, 20};
int k = 4;
int n = arr.length;
System.out.println(maxSum(arr, n, k));
}
}
// This code is contributed by Arnav Kr. Mandal.
```

Output :
`24`

Time Complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Abhishek Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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