I am not able to figure out my mistake

#1
#2

hello @vageesha
ur recurrence relation is wrong both playeers are playing optimally.
so ur opponnent step will also work affect ur score.
ply refer this->
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.

ll optimalGame(ll i,ll j){
    if(i > j){
        return 0;
    }

    // Consider both the possibilities. You can pick either the first or the last coin.
    // Since the opponent plays optimally , we would get the minimum of the remaining coins for each choice.
    ll pickFirst = coins[i] + min( optimalGame(i+2,j) , optimalGame(i+1,j-1) ) ;
    ll pickLast = coins[j] + min( optimalGame(i,j-2) , optimalGame(i+1,j-1) ) ;

    // Pick the max of two as your final result
    ll ans = max(pickFirst,pickLast);

    return ans;
}
#3

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