Skip to content
Related Articles
Open in App
Not now

Related Articles

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

Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 30 Jan, 2023
Improve Article
Save Article

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 <bits/stdc++.h>
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
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 <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 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 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.

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




<?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.
?>

Javascript




<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>

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 review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!