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

08 Jul, 2022
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
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. 
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).
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++
Java
Python3
C#
PHP
Javascript
// 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 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<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 codeint 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 Code for Find maximum (or minimum)// sum of a subarray of size kimport 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<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.
# 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 codearr = [1, 4, 2, 10, 2, 3, 1, 0, 20]k = 4n = len(arr)print(maxSum(arr, n, k)) # This code is contributed by Anant Agarwal.
// C# Code for Find maximum (or minimum)// sum of a subarray of size kusing 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// O(n) solution for finding maximum// sum of a subarray of size k // Returns maximum sum in a// subarray of size k.function maxSum($arr, $n, $k){         // k must be smaller than n    if ($n < $k)    {        echo "Invalid";        return -1;    }     // Compute sum of first    // window of size k    $res = 0;    for($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.    $curr_sum = $res;    for($i = $k; $i < $n; $i++)    {        $curr_sum += $arr[$i] -                     $arr[$i - $k];        $res = max($res, $curr_sum);    }     return $res;}     // Driver Code    $arr = array(1, 4, 2, 10, 2, 3, 1, 0, 20);    $k = 4;    $n = sizeof($arr);    echo maxSum($arr, $n, $k); // This code is contributed by nitin mittal.?>
<script>    // JAVA SCRIPT Code for Find maximum (or minimum)    // sum of a subarray of size k          // Returns maximum sum in a subarray of size k.    function maxSum(arr,n,k)    {        // k must be smaller than n        if (n < k)        {            document.write("Invalid");            return -1;        }             // Compute sum of first window of size k        let res = 0;        for (let 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.        let curr_sum = res;        for (let 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 */             let arr = [1, 4, 2, 10, 2, 3, 1, 0, 20];        let k = 4;        let n = arr.length;        document.write(maxSum(arr, n, k));     // This code is contributed by sravan kumar Gottumukkala</script>
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Time Complexity: O(n)Auxiliary Space: O(1) 
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