# 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 the sum of previous subarray (or window) of size k. Except for the first subarray of size k, for other subarrays, we compute sum by removing the first element of the last window and adding the last element of the current window.

Below is implementation of above idea.

## C++

 `// O(n) solution for finding maximum sum of ` `// a subarray of size k ` `#include ` `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

## 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

## Python3

 `# O(n) solution in Python3 for finding  ` `# maximum sum of a subarray of size k ` ` `  `# Returns maximum sum in ` `# a subarray of size k. ` `def` `maxSum(arr, n, k): ` ` `  `    ``# k must be greater ` `    ``if` `(n < k): ` `     `  `        ``print``(``"Invalid"``) ` `        ``return` `-``1` `     `  `    ``# Compute sum of first ` `    ``# window of size k ` `    ``res ``=` `0` `    ``for` `i ``in` `range``(k): ` `        ``res ``+``=` `arr[i] ` ` `  `    ``# Compute sums of remaining windows by ` `    ``# removing first element of previous ` `    ``# window and adding last element of  ` `    ``# current window. ` `    ``curr_sum ``=` `res ` `    ``for` `i ``in` `range``(k, n): ` `     `  `        ``curr_sum ``+``=` `arr[i] ``-` `arr[i``-``k] ` `        ``res ``=` `max``(res, curr_sum) ` ` `  `    ``return` `res ` ` `  `# Driver code ` `arr ``=` `[``1``, ``4``, ``2``, ``10``, ``2``, ``3``, ``1``, ``0``, ``20``] ` `k ``=` `4` `n ``=` `len``(arr) ` `print``(maxSum(arr, n, k)) ` ` `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# Code for Find maximum (or minimum)  ` `// sum of a subarray of size k ` `using` `System; ` ` `  `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) ` `        ``{ ` `            ``Console.Write(``"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 Code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {1, 4, 2, 10, 2, 3, 1, 0, 20}; ` `        ``int` `k = 4; ` `        ``int` `n = arr.Length; ` `        ``Console.Write(maxSum(arr, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` `

Output :

`24`

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

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Improved By : nitin mittal, tapas139

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