Maximum sum subarray of even length
Last Updated :
12 Jul, 2025
Given an array arr[] of N elements, the task is to find the maximum sum of any subarray of length X such that X > 0 and X % 2 = 0.
Examples:
Input: arr[] = {1, 2, 3}
Output: 5
{2, 3} is the required subarray.
Input: arr[] = {8, 9, -8, 9, 10}
Output: 20
{9, -8, 9, 10} is the required subarray.
Even though {8, 9, -8, 9, 10} has the maximum sum
but it is not of even length.
Approach: This problem is a variation of maximum subarray sum problem and can be solved using dynamic programming approach. Create an array dp[] where dp[i] will store the maximum sum of an even length subarray whose first element is arr[i]. Now the recurrence relation will be:
dp[i] = max((arr[i] + arr[i + 1]), (arr[i] + arr[i + 1] + dp[i + 2]))
This is because the maximum sum even length subarray starting with the element arr[i] can either be the sum of arr[i] and arr[i + 1] or it can be arr[i] + arr[i + 1] added with the maximum sum of even length subarray starting with arr[i + 2] i.e. dp[i + 2]. Take the maximum of these two.
In the end, the maximum value from the dp[] array will be the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum
// subarray sum of even length
int maxEvenLenSum(int arr[], int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int dp[n] = { 0 };
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--) {
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = *max_element(dp, dp + n);
return maxSum;
}
// Driver code
int main()
{
int arr[] = { 8, 9, -8, 9, 10 };
int n = sizeof(arr) / sizeof(int);
cout << maxEvenLenSum(arr, n);
return 0;
}
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to return the maximum
// subarray sum of even length
static int maxEvenLenSum(int arr[], int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int []dp = new int[n];
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--)
{
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = Arrays.stream(dp).max().getAsInt();
return maxSum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 9, -8, 9, 10 };
int n = arr.length;
System.out.println(maxEvenLenSum(arr, n));
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation of the approach
# Function to return the maximum
# subarray sum of even length
def maxEvenLenSum(arr, n):
# There has to be at
# least 2 elements
if (n < 2):
return 0
# dp[i] will store the maximum
# subarray sum of even length
# starting at arr[i]
dp = [0 for i in range(n)]
# Valid subarray cannot start from
# the last element as its
# length has to be even
dp[n - 1] = 0
dp[n - 2] = arr[n - 2] + arr[n - 1]
for i in range(n - 3, -1, -1):
# arr[i] and arr[i + 1] can be added
# to get an even length subarray
# starting at arr[i]
dp[i] = arr[i] + arr[i + 1]
# If the sum of the valid subarray
# starting from arr[i + 2] is
# greater than 0 then it can be added
# with arr[i] and arr[i + 1]
# to maximize the sum of the
# subarray starting from arr[i]
if (dp[i + 2] > 0):
dp[i] += dp[i + 2]
# Get the sum of the even length
# subarray with maximum sum
maxSum = max(dp)
return maxSum
# Driver code
arr = [8, 9, -8, 9, 10]
n = len(arr)
print(maxEvenLenSum(arr, n))
# This code is contributed by Mohit Kumar
// C# implementation of the approach
using System;
class GFG
{
static int MaxSum(int []arr)
{
// assigning first element to the array
int large = arr[0];
// loop to compare value of large
// with other elements
for (int i = 1; i < arr.Length; i++)
{
// if large is smaller than other element
// assign that element to the large
if (large < arr[i])
large = arr[i];
}
return large;
}
// Function to return the maximum
// subarray sum of even length
static int maxEvenLenSum(int []arr, int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
int []dp = new int[n];
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (int i = n - 3; i >= 0; i--)
{
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
int maxSum = MaxSum(dp);
return maxSum;
}
// Driver code
public static void Main()
{
int []arr = { 8, 9, -8, 9, 10 };
int n = arr.Length;
Console.WriteLine(maxEvenLenSum(arr, n));
}
}
// This code is contributed by kanugargng
<script>
// Javascript implementation of the approach
// Function to return the maximum
// subarray sum of even length
function maxEvenLenSum(arr, n) {
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// dp[i] will store the maximum
// subarray sum of even length
// starting at arr[i]
let dp = new Array(n).fill(0);
// Valid subarray cannot start from
// the last element as its
// length has to be even
dp[n - 1] = 0;
dp[n - 2] = arr[n - 2] + arr[n - 1];
for (let i = n - 3; i >= 0; i--) {
// arr[i] and arr[i + 1] can be added
// to get an even length subarray
// starting at arr[i]
dp[i] = arr[i] + arr[i + 1];
// If the sum of the valid subarray starting
// from arr[i + 2] is greater than 0 then it
// can be added with arr[i] and arr[i + 1]
// to maximize the sum of the subarray
// starting from arr[i]
if (dp[i + 2] > 0)
dp[i] += dp[i + 2];
}
// Get the sum of the even length
// subarray with maximum sum
let maxSum = dp.sort((a, b) => b - a)[0];
return maxSum;
}
// Driver code
let arr = [8, 9, -8, 9, 10];
let n = arr.length;
document.write(maxEvenLenSum(arr, n));
// This code is contributed by _saurabh_jaiswal.
</script>
Time complexity: O(n)
Space complexity: O(n)
Efficient approach : Space optimization O(1)
To optimize the space complexity of previous approach we using only two variables to keep track of the previous two subproblems instead of creating an array of size n to store all the subproblem solutions. This way, we can reduce the space complexity from O(n) to O(1).
Implementation Steps:
- Check if the array size is less than 2, return 0 if true.
- Initialize prevPrevSum to 0 and prevSum to arr[n-2] + arr[n-1], which represents the sum of the last two elements of the array.
- Traverse the array from the second-last index to the first index and compute the sum of every even length subarray that ends at the current index.
- Add the current element and the next element of the array to get the sum of the current even-length subarray.
- If the sum of the subarray two indices ahead is greater than 0, add it to the current subarray sum.
- Update prevPrevSum and prevSum with the previous subarray sums for further iterations.
- At last Return the maximum of prevSum and prevPrevSum.
Implementation:
C++
// C++ code for above approach
#include <bits/stdc++.h>
using namespace std;
int maxEvenLenSum(int arr[], int n)
{
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// initialize variables to store the previous values
int prevPrevSum = 0, prevSum = arr[n - 2] + arr[n - 1], currSum;
// iterate over subproblems and get the current value from previous computations
for (int i = n - 3; i >= 0; i--) {
currSum = arr[i] + arr[i + 1];
if (prevPrevSum > 0)
currSum += prevPrevSum;
// assigning values for further iterations
prevPrevSum = prevSum;
prevSum = currSum;
}
// return answer
return max(prevSum, prevPrevSum);
}
// Driver code
int main()
{
int arr[] = { 8, 9, -8, 9, 10 };
int n = sizeof(arr) / sizeof(int);
// function call
cout << maxEvenLenSum(arr, n);
return 0;
}
// Java code for above approach
import java.util.*;
public class Main {
static int maxEvenLenSum(int arr[], int n) {
// There has to be at
// least 2 elements
if (n < 2)
return 0;
// initialize variables to store the previous values
int prevPrevSum = 0, prevSum = arr[n - 2] + arr[n - 1], currSum;
// iterate over subproblems and get the current value from previous computations
for (int i = n - 3; i >= 0; i--) {
currSum = arr[i] + arr[i + 1];
if (prevPrevSum > 0)
currSum += prevPrevSum;
// assigning values for further iterations
prevPrevSum = prevSum;
prevSum = currSum;
}
// return answer
return Math.max(prevSum, prevPrevSum);
}
// Driver code
public static void main(String[] args) {
int arr[] = { 8, 9, -8, 9, 10 };
int n = arr.length;
// function call
System.out.println(maxEvenLenSum(arr, n));
}
}
def maxEvenLenSum(arr, n):
# There has to be at least 2 elements
if n < 2:
return 0
# initialize variables to store the previous values
prevPrevSum = 0
prevSum = arr[n - 2] + arr[n - 1]
# iterate over subproblems and get the current value from previous computations
for i in range(n - 3, -1, -1):
currSum = arr[i] + arr[i + 1]
if prevPrevSum > 0:
currSum += prevPrevSum
# assigning values for further iterations
prevPrevSum = prevSum
prevSum = currSum
# return answer
return max(prevSum, prevPrevSum)
# Driver code
arr = [8, 9, -8, 9, 10]
n = len(arr)
# function call
print(maxEvenLenSum(arr, n))
using System;
class GFG {
static int maxEvenLenSum(int[] arr, int n)
{
// There has to be at least 2 elements
if (n < 2)
return 0;
// initialize variables to store the previous values
int prevPrevSum = 0,
prevSum = arr[n - 2] + arr[n - 1], currSum;
// iterate over subproblems and get the current
// value from previous computations
for (int i = n - 3; i >= 0; i--) {
currSum = arr[i] + arr[i + 1];
if (prevPrevSum > 0)
currSum += prevPrevSum;
// assigning values for further iterations
prevPrevSum = prevSum;
prevSum = currSum;
}
// return answer
return Math.Max(prevSum, prevPrevSum);
}
// Driver code
static void Main()
{
int[] arr = { 8, 9, -8, 9, 10 };
int n = arr.Length;
// function call
Console.Write(maxEvenLenSum(arr, n));
}
}
<script>
// Javascript code for above approach
function maxEvenLenSum(arr, n) {
// There has to be at
// least 2 elements
if (n < 2) {
return 0;
}
// Initialize variables to store the previous values
let prevPrevSum = 0;
let prevSum = arr[n - 2] + arr[n - 1];
let currSum;
// Iterate over subproblems and get the current value from previous computations
for (let i = n - 3; i >= 0; i--) {
currSum = arr[i] + arr[i + 1];
if (prevPrevSum > 0) {
currSum += prevPrevSum;
}
// Assign values for further iterations
prevPrevSum = prevSum;
prevSum = currSum;
}
// Return answer
return Math.max(prevSum, prevPrevSum);
}
// Driver code
const arr = [8, 9, -8, 9, 10];
const n = arr.length;
// Function call
document.write(maxEvenLenSum(arr, n));
// This code is contributed by Vaibhav Nandan
</script>
Output
20
Time complexity: O(n)
Auxiliary Space: O(1)
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