HackerRank Bob and Ben problem solution

In this HackerRank Bob and Ben problem solution we have given the values of Mi the number of nodes in the tree and a constant Ki for each tree in the forest, we need to determine who will win the game.

HackerRank Bob and Ben problem solution in Python.

#!/bin/python3

import os
import sys
import functools

#
# Complete the bobAndBen function below.
#
def bobAndBen(trees):
    #
    # Write your code here.
    #
    nims = []
    for tree in trees:
        nims.append(grundy(tree[0]))
    res = functools.reduce(lambda a,b: a^b, nims)
    if res == 0: return "BEN"
    return "BOB"

def grundy(m):
    res = 0 if m == 0 or m == 2 else ((m-1) % 2) + 1
    return res

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    g = int(input())

    for g_itr in range(g):
        n = int(input())

        trees = []

        for _ in range(n):
            trees.append(list(map(int, input().rstrip().split())))

        result = bobAndBen(trees)

        fptr.write(result + 'n')

    fptr.close()

Bob and Ben problem solution in Java.

import java.io.*;
import java.util.*;
public class Solution
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        int G = input.nextInt();
        for(int g=0; g<G; g++)
        {
            int n = input.nextInt();
            int p = 0;
            for(int i=0; i<n; i++)
            {
                int m = input.nextInt();
                int k = input.nextInt();
                if(m%2 == 1)
                    p = p^1;
                else if(m > 2)
                    p = p^2;
            }
            if(p==0)
                System.out.println("BEN");
            else
                System.out.println("BOB");
        }
    }
}

Problem solution in C++.

#include<string>
#include<iostream>
#include<fstream>
#include<stdio.h>
#include<string.h>
#include<memory.h>
#include<math.h>
#include<vector>
#include<algorithm>
#include<map>
#include<numeric>
#include<deque>
#include<set>
#include<functional>
#include<queue>
#include<stack>
#include<unordered_set>
#include<unordered_map>



#define REP(i,s,n) for(int (i)=s; (i)<(int)(n);(i)++)
#define RIT(it,c) for(__typeof(c.begin()) it = c.begin();it!=c.end();it++)
#define RITT(it,v,c) for(__typeof(c.begin()) it = v;it!=c.end();it++)
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int)(x).size()
#define MSET(m,v) memset(m,v,sizeof(m))
#define auto(c) __typeof(c.begin())

using namespace std;
typedef vector<int> vi;
typedef pair<int,int>ii;
typedef long long LL;
typedef pair<ii,int> iiit;



int main(){
    int T;
    cin>>T;
    for(int k=0;k<T;k++){
        int n,sum=0;
        cin>>n;
        int cnt = n;
        for(int i=0;i<n;i++){
            int m,l;
            scanf("%d%d",&m,&l);
            if(m==2) cnt++;
            sum += m%2;
        }
        if((cnt%2==0 && sum%2==0)) cout<<"BEN"<<endl;
        else cout<<"BOB"<<endl;
    }
}

Problem solution in C.

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>

int main()
{
    int test,t;
    scanf("%d",&test);
    for(t=1;t<=test;t++)
    {
        int n,i;
        scanf("%d",&n);
        int answer=0;
        for(i=0;i<n;i++)
        {
            int m,k;
            scanf("%d %d",&m,&k);
            if(m==2)
            {
                m=0;
            }
            else if(m==3)
            {
                m=1;
            }
            else if(m>3)
            {
                if(m%2==0)
                {
                    m=2;
                }
                else
                {
                    m=1;
                }
            }
            answer=answer^m;
        }
        if(answer==0)
        {
            printf("BENn");
        }
        else
        {
            printf("BOBn");
        }
    }
    return 0;
}