Find maximum (or minimum) sum of a subarray of size k
Last Updated :
23 Jul, 2025
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 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
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# program to demonstrate deletion in
// Ternary Search Tree (TST)
// For insert and other functions, refer
// https://www.geeksforgeeks.org/dsa/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.
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.
OutputMax Sum By Brute Force :24
Time Complexity: O(n2)
Auxiliary Space: O(1)
Using Queue:
We can use queue structure to calculate max or min sum of a subarray of size k.
Algorithm:
- First create an queue structure and push k elements inside it and calculate the sum of the elements (let's say su) during pushing.
- Now create a max/min variable (let's say m) with value INT_MIN for max value or INT_MAX for min value.
- Then iterate using loop from kth position to the end of the array.
- During each iteration do below steps:
- First subtract su with the front element of the queue.
- Then pop it from the queue.
- Now, push the current element of the array inside queue and add it to the su.
- Then compare it with the value m for max/min.
- Now, print the current m value.
Below is the implementation of the above approach:
C++
// O(n) solution for finding maximum sum of
// a subarray of size k with O(k) space
#include <bits/stdc++.h>
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;
}
// Create Queue
queue<int> q;
int m = INT_MIN;
int su = 0;
// Compute sum of first k elements and also
// store then inside queue
for (int i = 0; i < k; i++) {
q.push(arr[i]);
su += arr[i];
}
for (int i = k; i < n; i++) {
m = max(m, su);
su -= q.front();
q.pop();
q.push(arr[i]);
su += arr[i];
}
m = max(m, su);
return m;
}
// 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;
}
// This code is contributed by Susobhan Akhuli
// Java equivalent code
import java.util.LinkedList;
import java.util.Queue;
class GFG {
// Function to find maximum sum of
// a subarray of size k
static int maxSum(int arr[], int n, int k)
{
// k must be smaller than n
if (n < k) {
System.out.println("Invalid");
return -1;
}
// Create Queue
Queue<Integer> q = new LinkedList<Integer>();
// Initialize maximum and current sum
int m = Integer.MIN_VALUE;
int su = 0;
// Compute sum of first k elements
// and also store them inside queue
for (int i = 0; i < k; i++) {
q.add(arr[i]);
su += arr[i];
}
// Compute sum of remaining elements
for (int i = k; i < n; i++) {
m = Math.max(m, su);
// remove first element from the queue
su -= q.peek();
q.remove();
// add current element to the queue
q.add(arr[i]);
su += arr[i];
// update maximum sum
}
m = Math.max(m, su);
return m;
}
// Driver code
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 Susobhan Akhuli
from collections import deque
def maxSum(arr, n, k):
# k must be smaller than n
if n < k:
print("Invalid")
return -1
# Create deque
q = deque()
# Initialize maximum and current sum
m = float('-inf')
su = 0
# Compute sum of first k elements
# and also store them inside deque
for i in range(k):
q.append(arr[i])
su += arr[i]
# Compute sum of remaining elements
for i in range(k, n):
m = max(m, su)
# Remove first element from the deque
su -= q[0]
q.popleft()
# Add current element to the deque
q.append(arr[i])
su += arr[i]
# Update maximum sum for the last window
m = max(m, su)
return m
# Driver code
arr = [ 1, 4, 2, 10, 2, 3, 1, 0, 20]
k = 4
n = len(arr)
print(maxSum(arr, n, k))
// O(n) solution for finding maximum sum of
// a subarray of size k with O(k) space
using System;
using System.Collections.Generic;
class MainClass {
// Returns maximum sum in a subarray of size k.
static int maxSum(int[] arr, int n, int k)
{
// k must be smaller than n
if (n < k) {
Console.WriteLine("Invalid");
return -1;
}
// Create Queue
Queue<int> q = new Queue<int>();
int m = int.MinValue;
int su = 0;
// Compute sum of first k elements and also
// store then inside queue
for (int i = 0; i < k; i++) {
q.Enqueue(arr[i]);
su += arr[i];
}
for (int i = k; i < n; i++) {
m = Math.Max(m, su);
su -= q.Peek();
q.Dequeue();
q.Enqueue(arr[i]);
su += arr[i];
}
m = Math.Max(m, su);
return m;
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 1, 4, 2, 10, 2, 3, 1, 0, 20 };
int k = 4;
int n = arr.Length;
Console.WriteLine(maxSum(arr, n, k));
}
}
// This code is contributed by Susobhan Akhuli
// JavaScript solution for finding maximum sum of
// a subarray of size k with O(k) space
// Returns maximum sum in a subarray of size k.
function maxSum(arr, n, k)
{
// k must be smaller than n
if (n < k) {
console.log("Invalid");
return -1;
}
// Create Queue
const q = [];
let m = Number.MIN_SAFE_INTEGER;
let su = 0;
// Compute sum of first k elements and also
// store then inside queue
for (let i = 0; i < k; i++) {
q.push(arr[i]);
su += arr[i];
}
for (let i = k; i < n; i++) {
m = Math.max(m, su);
su -= q.shift();
q.push(arr[i]);
su += arr[i];
}
m = Math.max(m, su);
return m;
}
// Driver code
const arr = [ 1, 4, 2, 10, 2, 3, 1, 0, 20 ];
const k = 4;
const n = arr.length;
console.log(maxSum(arr, n, k));
// This code is contributed by Susobhan Akhuli.
Time Complexity: O(n), to iterate n times.
Auxiliary Space: O(k) , to store k elements inside queue.
Optimized Solution (Space Optimization) :
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 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.
# 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# 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.
<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>
<?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.
?>
Time Complexity: O(n)
Auxiliary Space: O(1)
PROBLEM OF THE DAY : 11/12/2023 | Max Sum Subarray of size K
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