Sir, i'm unable to implement the another part of code

#1

sir, how can i implement the alternate turn for ‘‘nimit’’. here is my code -> https://ide.codingblocks.com/s/458942

#2

hello @chandreshmaurya
you can pass a bool variable to indicate the turn (0 for nimit, 1 piyush).
after the turn is complete and when u make next call then flip the turn value (ie if 0 then 1 ,if 1 then 0).

also greedy approach will not work here. try to think of another approach or just try to simulate the situation using recusion.

#3

sir, i got the concept:- max((a[s] + min((s+2 —>e), (s+1—>e-1))), (a[e] + min((s+1—>e-1), (s—>e-2))))

#4

but, unable to implement that

#5

check this->

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;
}
#6

please help me to implement this apporoach.

#7

…

check my last response

#8

yes sir, i,m going through that

#9

sir, i code it with the help of u and got correct answer. but, i have a doubt in the line of 7, what does the mean of “return 0;” if (s>e). here is my code --> https://ide.codingblocks.com/s/459019

#10

it is the case when we have no element left in the array and simply in that case max score possible is 0, that is why we are returning 0

#11

ok sir, i got it. thank you sir :heart:

#12

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