Skip to content
Related Articles

Related Articles

Find maximum (or minimum) sum of a subarray of size k
  • Difficulty Level : Easy
  • Last Updated : 09 May, 2020

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.

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 <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.


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




<?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 greater
    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.
?>



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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :