I need some hint for corner cases

my code fails when input is of the form
2
3
2
23
2332
32
2
because of same values of top and bottom coins, i know that at this condition we need to make the decision on the basis of the inner placement of coins but i can’t code it

we would need an optimal solution for this. At each instance we would need to consider two possibilities that we can pick the first as well as the last element of the remaining array. Both these possibilities give rise to two more possibilities depending on the other player. Since the second player plays optimally and try to minimise our score. So overall we have two possibilities at each instance.
For the first possibility , where we could pick the first element , the other player will pick the next element from the side that would minimise our total score.
Similarly , for the second possibility , where we can pick the last element , the other player would still pick the next element from the side that would minimise our total score.
We entertain both these cases and take the maximum result of the two and return that result.
We take two pointer variables , say ‘i’ and ‘j’ which each represent the starting and the ending point of the remaining array currently in consideration. We work till the two pointers cross each other.
see this: