In this Counter game HackerRank solution, Louise and Richard have developed a numbers game. They pick a number and check to see if it is a power of . If it is, they divide it by . If not, they reduce it by the next lower number which is a power of . Whoever reduces the number to wins the game. Louise always starts.
Given an initial value, determine who wins the game.
Example
It’s Louise’s turn first. She determines that is not a power of . The next lower power of is , so she subtracts that from and passes to Richard. is a power of , so Richard divides it by and passes to Louise. Likewise, is a power so she divides it by and reaches . She wins the game.
Update If they initially set counter to , Richard wins. Louise cannot make a move so she loses.
Function Description
Complete the counterGame function in the editor below.
counterGame has the following parameter(s):
- int n: the initial game counter value
Returns
- string: either
RichardorLouise
Input Format
The first line contains an integer , the number of testcases.
Each of the next lines contains an integer , the initial value for each game.
Constraints
Sample Input
1
6
Sample Output
Richard
Explanation
- As is not a power of , Louise reduces the largest power of less than i.e., , and hence the counter reduces to .
- As is a power of , Richard reduces the counter by half of i.e., . Hence the counter reduces to .
As we reach the terminating condition with , Richard wins the game.
Counter game HackerRank solution
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in C#
public static string counterGame(long n)
{
bool louiseWins = false;
while (n>1)
{
var log = Math.Log2(n);
if ((log % 1) != 0)
{
log = Math.Floor(log);
var next = (long)Math.Pow(2, log);
n = (n - next);
}
else
{
n = n / 2;
}
louiseWins = !louiseWins;
}
return louiseWins ? "Louise" : "Richard";
}Problem Solution in Python
def counterGame(n):
louise = False
while n > 1:
if n & (n-1) == 0:
# is power of 2
n = n >> 1
louise = not louise
else:
x = 1
while x < n:
x = x << 1 # next power of 2
x = x >> 1
n = n - x
louise = not louise
return 'Louise' if louise else 'Richard'Problem Solution in JavaScript
function counterGame(n) {
let turn = 0;
while (n > 1) {
turn++;
// if is power of 2, divide by 2
if ((n & (n - 1)) === 0) n >>>= 1;
// else subtract next lowest power of 2
else {
let p = n;
while ((p & (p - 1)) !== 0) {
p = p & (p - 1);
}
n -= p;
}
}
return turn % 2 === 0 ? 'Richard' : 'Louise';
}Problem Solution in C++
string counterGame(long n) {
bool player = true;
while(fabs(floor(log2(n)) - log2(n)) > 0.001)
{
n = n - pow(2, floor(log2(n)));
player = !player;
}
if (int(log2(n)) % 2 == 1)
player = !player;
return player ? "Richard" : "Louise";
}