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  1. Prepare
  2. Algorithms
  3. Game Theory
  4. Bob and Ben
  5. Discussions

Bob and Ben

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9 Discussions

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  • ffadvanceserver1
     + 0 comments

    This was a fun one — XOR problems always force you to think a bit differently. I’ve been into debugging game mechanics lately too, especially around how logic plays out in advance servers. Challenges like this help sharpen that mindset a lot.

  • thecodingsoluti2
     + 0 comments

    Here is Bob and Ben problem solution in Python Java C++ and c programming - https://programs.programmingoneonone.com/2021/07/hackerrank-bob-and-ben-problem-solution.html

  • ygrinev
     + 0 comments

    C# (as usually short - 1 line - and elegant, no indexing):

    public static string bobAndBen(List<List<int>> trees)
{
    return trees.Aggregate(0,(xor,t)=>xor^(t[0]==0 || t[0]==2 ? 0 : (t[0]-1)%2+1))==0
    ? "BEN" : "BOB";
}

    Python must be 1-ne as well........

  • xdavidliu
     + 0 comments

    simple java solution, O(1) space and O(number of trees) time

    import java.util.Scanner;

class Solution{
    static void solve(){
	Scanner sc=new Scanner(System.in);
	int ngame=sc.nextInt();
	for(int i=0;i<ngame;++i){
	    int ntree=sc.nextInt(), xor=0;
	    for(int j=0;j<ntree;++j){
		int nnode=sc.nextInt();
		sc.nextInt();
		int nimber=0;
	        if(nnode!=2) nimber=2-(nnode&1);
		xor=(j==0?nimber: xor^nimber);
	    }
	    System.out.println(xor==0?"BEN":"BOB");
	}
	sc.close();
    }
    public static void main(String[] args){
	solve();
    }
}

  • kurimanish049
     + 1 comment
    def GrudyNumber(nodes):
    if(nodes == 0 or nodes == 2):
        return 0 
    else:
        return ((nodes-1)%2+1)

def bobAndBen(trees): 
    GrudyNumers= []
    for i in range(len(trees)):
        GrudyNumers.append(GrudyNumber(trees[i][0]))
    re=GrudyNumers[0]
    print(GrudyNumers)
    if((len(GrudyNumers)-1)>1):
        for j in range(1,len(GrudyNumers)):
            re^=GrudyNumers[j]
    elif((len(GrudyNumers)-1)==1):
        re=re ^ GrudyNumers[1]
    if(re != 0): 
        return 'BOB' 
    else:
        return 'BEN'