Maximize Score

/* given array, perform k operations:
in one operation: choose an element from only end(front or back) -> add to score and remove from array.
return max score, can be obtained this way.

Sample Input 1:
2
5 1
10 2 8 9 2
5 2
10 5 6 7 9
Sample Output 1:
10 
19
Explanation Of Sample Input 1:
Test Case 1 :  
Given N = 5 and K = 1.
arr = [10, 2, 8, 9, 2]
Initial score = 0
In the first operation choose 10 from the start and add it to the score.
So final score = 10.

Test Case 2 : 
Given N = 5 and K = 2.
arr = [10, 5, 6, 7, 9]
Initial score = 0
In the first operation choose 10 from the start and add it to the score.
So score = 10 and arr = [5, 6, 7, 9]
In the second operation choose 9 from the end and add it to the score.
So final score = 10 + 9 = 19.
Sample Input 2:
2
4 4
10 2 2 2
4 3
10 1 7 9
Sample Output 2:
16
26
Explanation Of Sample Input 2:
Test Case 1 :  
Given N = 4 and K = 4.
arr = [10, 2, 2, 2]
Initial score = 0
Here  N = K. So we can add all elements to our score by always picking array elements from the start.
So final score = 10 + 2 + 2 + 2 = 16.

Test Case 2 : 
Given N = 4 and K = 3.
arr = [10, 1, 7, 9]
Initial score = 0
In the first operation choose 10 from the start and add it to the score.
So score = 10 and arr = [1, 7, 9]
In the second operation choose 9 from the end and add it to the score.
So score = 10 + 9 = 19 and and arr = [1, 7]
In the third operation choose 7 from the end and add it to the score.
So final score = 19 + 7 = 26.
*/

inline int maximizeScore(vector<int> &arr, int n, int k)
{
    int ksum = 0;
    for(int i=0;i<k;++i) ksum += arr[i];
    int ans = ksum;
    for(int i=n-1;i>=0 && k; --i,--k){
        ksum += arr[i] - arr[k-1];
        ans = max(ans, ksum);
    }
    return ans;
}