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  1. Counter game
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Counter game

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210 Discussions

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  • fazepavan
     + 0 comments
    def counterGame(n):
    # Write your code here
    turns = 0
    while n > 1:
        if(n&(n-1) == 0):
            n//=2
        else:
            highest_power_of_2 = 1
            while highest_power_of_2*2<=n:
                highest_power_of_2*=2
            n-=highest_power_of_2
        turns += 1
    if turns %2 == 0:
        return "Richard"
    else:
        return "Louise"

  • trinm1308
     + 0 comments

    Javascript

    function counterGame(n) {
    // Write your code here    
    var turn = 0
    while(n > 1){
        n = Math.log2(n) % 1 == 0 ? n/2 : n - Math.pow(2, Math.floor(Math.log2(n)))        
        turn ++
    }
    return turn % 2 == 0 ? 'Richard' : 'Louise'
}

  • garg_jyoti
     + 0 comments

    Java solution class Result {

    public static String counterGame(long n) {
    if (n == 1) return "Richard";

    int turn = 0;
    while (n > 1) {
        if ((n & (n - 1)) == 0) { // Check if n is a power of two
            n /= 2;
        } else {
            n -= Long.highestOneBit(n); // Get the largest power of 2 less than or equal to n
        }
        turn++;
    }
    return (turn % 2 == 0) ? "Richard" : "Louise";
}

    }

  • nobodys_ghost
     + 0 comments

    Javascript 

    function counterGame(n) {
    let turn = 0;
    let log = Math.log2(n);
    while (log % 1 !== 0) {
        n -= Math.pow(2, Math.floor(log)); //subtract nearest power of 2
        turn++;
        log = Math.log2(n);
    }
    const endTurn = turn + log;
    return endTurn % 2 === 0 ? "Richard" : "Louise";
}

  • mcoroz96
     + 0 comments

    c++

    string counterGame(long n) {

 std::bitset<sizeof(long) * 8> bits = n-1;
 
 if(bits.count() % 2 == 0){
    return "Richard";
 }
 return "Louise";
}