Optimal Game Strategy-1

#1

What is the problem in this code??

#2

try for this input
4
1 99 1001 100
here answer is: 1002
This is a classic dp problem.

#3

dp has not been covered in the course yet.
i just completed recursion.
so what should i do?

#4

Constraints are small, try using recursion only.

#5

i am not able to solve it. please help

#6

try brute i, try all possiblities.
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.

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;
}
#7

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