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

• Difficulty Level : Easy
• Last Updated : 30 Jan, 2023

Given an array of integers and a number k, find the 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
Explanation: We get maximum sum by adding subarray {4, 2, 10, 23} of size 4.

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

Naive Solution :

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

Below is the implementation of the above idea.

## C++

 `#include ``using` `namespace` `std;``int` `max_sum_of_subarray(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `max_sum = 0;``    ``for` `(``int` `i = 0; i + k <= n; i++) {``        ``int` `temp = 0;``        ``for` `(``int` `j = i; j < i + k; j++) {``            ``temp += arr[j];``        ``}``        ``if` `(temp > max_sum)``            ``max_sum = temp;``    ``}` `    ``return` `max_sum;``}``int` `main()``{``    ``int` `arr[] = { 1, 4, 2, 10, 2, 3, 1, 0, 20 };``    ``int` `k = 4;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``    ``int` `max_sum;` `    ``// by brute force``    ``max_sum = max_sum_of_subarray(arr, n, k);``    ``cout << max_sum << endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;` `public` `class` `GFG {``    ``static` `int` `max_sum_of_subarray(``int` `arr[], ``int` `n, ``int` `k)``    ``{``        ``int` `max_sum = ``0``;``        ``for` `(``int` `i = ``0``; i + k <= n; i++) {``            ``int` `temp = ``0``;``            ``for` `(``int` `j = i; j < i + k; j++) {``                ``temp += arr[j];``            ``}``            ``if` `(temp > max_sum)``                ``max_sum = temp;``        ``}` `        ``return` `max_sum;``    ``}``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``4``, ``2``, ``10``, ``2``, ``3``, ``1``, ``0``, ``20` `};``        ``int` `k = ``4``;``        ``int` `n = arr.length;` `        ``// by brute force``        ``int` `max_sum = max_sum_of_subarray(arr, n, k);``        ``System.out.println(max_sum);``    ``}``}` `// This code is contributed by Karandeep1234`

## Python3

 `def` `max_sum_of_subarray(arr, n, k):``    ``max_sum ``=` `0``;``    ``for` `i ``in` `range``(``0``, n``-``k``+``1``):``        ``temp ``=` `0``;``        ``for` `j ``in` `range``(i, i``+``k):``            ``temp ``+``=` `arr[j];` `        ``if` `(temp > max_sum):``            ``max_sum ``=` `temp;` `    ``return` `max_sum;`  `arr ``=` `[ ``1``, ``4``, ``2``, ``10``, ``2``, ``3``, ``1``, ``0``, ``20` `];``k ``=` `4``;``n ``=` `len``(arr);``max_sum``=``0``;` ` ``# brute force``max_sum ``=` `max_sum_of_subarray(arr, n, k);``print``(max_sum);``   ` `  ``# This code is contributed by poojaagarwal2.`

## C#

 `// C# program to demonstrate deletion in``// Ternary Search Tree (TST)``// For insert and other functions, refer``// https://www.geeksforgeeks.org/ternary-search-tree``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;``using` `System.Text.RegularExpressions;` `public` `class` `Gfg``{``  ``static` `int` `max_sum_of_subarray(``int``[] arr, ``int` `n, ``int` `k)``  ``{``    ``int` `max_sum = 0;``    ``for` `(``int` `i = 0; i + k <= n; i++) {``      ``int` `temp = 0;``      ``for` `(``int` `j = i; j < i + k; j++) {``        ``temp += arr[j];``      ``}``      ``if` `(temp > max_sum)``        ``max_sum = temp;``    ``}` `    ``return` `max_sum;``  ``}``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] arr = { 1, 4, 2, 10, 2, 3, 1, 0, 20 };``    ``int` `k = 4;``    ``int` `n = arr.Length;``    ``int` `max_sum;` `    ``// by brute force``    ``max_sum = max_sum_of_subarray(arr, n, k);``    ``Console.Write(max_sum);``  ``}``}` `// This code is contributed by agrawalpoojaa976.`

## Javascript

 `function` `max_sum_of_subarray(arr, n, k)``{``    ``let max_sum = 0;``    ``for` `(let i = 0; i + k <= n; i++) {``        ``let temp = 0;``        ``for` `(let j = i; j < i + k; j++) {``            ``temp += arr[j];``        ``}``        ``if` `(temp > max_sum)``            ``max_sum = temp;``    ``}` `    ``return` `max_sum;``}` `let arr = [ 1, 4, 2, 10, 2, 3, 1, 0, 20 ];``let k = 4;``let n = arr.length;``let max_sum;` `// by brute force``max_sum = max_sum_of_subarray(arr, n, k);``console.log(max_sum);` `// This code is contributed by ritaagarwal.`

Output

`Max Sum By Brute Force :24`

Time Complexity: O(n2)
Auxiliary Space: O(1)

Optimized  Solution :

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

Below is the implementation of the 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 smaller than n``    ``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 smaller than n``        ``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 smaller than n``    ``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 smaller than n``        ``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

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## Javascript

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Output

`24`

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

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