First we only need to take subarrays with even size as it will change index parity. Now, we take sum of numbers in eve indexes and odd indexes and if ever sum in eve indexes go above odd indexes we can safely discard the subarray as it will only cause a decrease in the total sum. Now as every element has two neighbors except the boundary elements we need to check for both neighbors.

As we can exclude some subarrays from making our answer smaller, we would exclude the subarray with: sum of eve elements > sum of odd elements We can make an array which will contain: arr[2*i+1]-arr[2*i] that is, the difference of consecutive odd and even indexes. Now our problem boils down to, taking K subarrays such that sum is maximum.