Two players are playing a series of games of Rock–paper–scissors. There are a total of K games played. Player 1 has a sequence of moves denoted by string A and similarly player 2 has string B. If any player reaches the end of their string, they move back to the start of the string. The task is to count the number of games won by each of the player when exactly K games are being played.
Examples:
Input: k = 4, a = “SR”, b = “R”
Output: 0 2
Game 1: Player1 = S, Player2 = R, Winner = Player2
Game 2: Player1 = R, Player2 = R, Winner = Draw
Game 3: Player1 = S, Player2 = R, Winner = Player2
Game 4: Player1 = R, Player2 = R, Winner = DrawInput: k = 3, a = “S”, b = “SSS”
Output: 0 0
All the games are draws.
Approach: Let length of string a be n and length of string b be m. The observation here is that the games would repeat after n * m moves. So, we can simulate the process for n * m games and then count the number of times it gets repeated. For the remaining games, we can again simulate the process since it would be now smaller than n * m. For example, in the first example above, n = 2 and m = 1. So, the games will repeat after every n * m = 2 * 1 = 2 moves i.e. (Player2, Draw), (Player2, Draw), ….., (Player2, Draw).
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function that returns 1 if first player wins, // 0 in case of a draw and -1 if second player wins int compare( char first, char second)
{ // If both players have the same move
// then it's a draw
if (first == second)
return 0;
if (first == 'R' ) {
if (second == 'S' )
return 1;
else
return -1;
}
if (first == 'P' ) {
if (second == 'R' )
return 1;
else
return -1;
}
if (first == 'S' ) {
if (second == 'P' )
return 1;
else
return -1;
}
} // Function that returns the count of games // won by both the players pair< int , int > countWins( int k, string a, string b)
{ int n = a.length();
int m = b.length();
int i = 0, j = 0;
// Total distinct games that can be played
int moves = n * m;
pair< int , int > wins = { 0, 0 };
while (moves--) {
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins.first++;
// Player 2 wins the current game
if (res == -1)
wins.second++;
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins.first *= repeat;
wins.second *= repeat;
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem--) {
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins.first++;
// Player 2 wins the current game
if (res == -1)
wins.second++;
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
} // Driver code int main()
{ int k = 4;
string a = "SR" , b = "R" ;
auto wins = countWins(k, a, b);
cout << wins.first << " " << wins.second;
} |
// Java implementation of the above approach import java.util.*;
import java.awt.Point;
class GFG{
// Function that returns 1 if first player wins, // 0 in case of a draw and -1 if second player wins public static int compare( char first, char second)
{ // If both players have the same move
// then it's a draw
if (first == second)
return 0 ;
if (first == 'R' )
{
if (second == 'S' )
return 1 ;
else
return - 1 ;
}
if (first == 'P' )
{
if (second == 'R' )
return 1 ;
else
return - 1 ;
}
if (first == 'S' )
{
if (second == 'P' )
return 1 ;
else
return - 1 ;
}
return 0 ;
} // Function that returns the count of games // won by both the players public static Point countWins( int k, String a,
String b)
{ int n = a.length();
int m = b.length();
int i = 0 , j = 0 ;
// Total distinct games that
// can be played
int moves = n * m;
Point wins = new Point ( 0 , 0 );
while (moves-- > 0 )
{
int res = compare(a.charAt(i),
b.charAt(j));
// Player 1 wins the current game
if (res == 1 )
wins = new Point(wins.x + 1 ,
wins.y);
// Player 2 wins the current game
if (res == - 1 )
wins = new Point(wins.x,
wins.y + 1 );
i = (i + 1 ) % n;
j = (j + 1 ) % m;
}
// Number of times the above
// n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins = new Point(wins.x * repeat,
wins.y * repeat);
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem-- > 0 )
{
int res = compare(a.charAt(i),
b.charAt(j));
// Player 1 wins the current game
if (res == 1 )
wins = new Point(wins.x + 1 ,
wins.y);
// Player 2 wins the current game
if (res == - 1 )
wins = new Point(wins.x,
wins.y + 1 );
i = (i + 1 ) % n;
j = (j + 1 ) % m;
}
return wins;
} // Driver code public static void main(String[] args)
{ int k = 4 ;
String a = "SR" , b = "R" ;
Point wins = countWins(k, a, b);
System.out.println(wins.x + " " + wins.y);
} } // This code is contributed by divyeshrabadiya07 |
# Python3 implementation of the above approach # Function that returns 1 if first # player wins, 0 in case of a draw # and -1 if second player wins def compare(first, second):
# If both players have the same
# move then it's a draw
if (first = = second):
return 0
if (first = = 'R' ):
if (second = = 'S' ):
return 1
else :
return - 1
if (first = = 'P' ):
if (second = = 'R' ):
return 1
else :
return - 1
if (first = = 'S' ):
if (second = = 'P' ):
return 1
else :
return - 1
# Function that returns the count # of games won by both the players def countWins(k, a, b):
n = len (a)
m = len (b)
i = 0
j = 0
# Total distinct games that
# can be played
moves = n * m
wins = [ 0 , 0 ]
while (moves > 0 ):
res = compare(a[i], b[j])
# Player 1 wins the current game
if (res = = 1 ):
wins[ 0 ] + = 1
# Player 2 wins the current game
if (res = = - 1 ):
wins[ 1 ] + = 1
i = (i + 1 ) % n
j = (j + 1 ) % m
moves - = 1
# Number of times the above
# n * m games repeat
repeat = k / / (n * m)
# Update the games won
wins[ 0 ] * = repeat
wins[ 1 ] * = repeat
# Remaining number of games after
# removing repeated games
rem = k % (n * m)
while (rem > 0 ):
res = compare(a[i], b[j])
# Player 1 wins the current game
if (res = = 1 ):
wins[ 0 ] + = 1
# Player 2 wins the current game
if (res = = - 1 ):
wins[ 1 ] + = 1
i = (i + 1 ) % n
j = (j + 1 ) % m
return wins
# Driver code if __name__ = = "__main__" :
k = 4
a = "SR"
b = "R"
wins = countWins(k, a, b);
print (wins[ 0 ], wins[ 1 ])
# This code is contributed by chitranayal |
// C# implementation of the above approach using System;
using System.Collections.Generic;
class GFG{
// Function that returns 1 if first player // wins, 0 in case of a draw and -1 if // second player wins static int compare( char first, char second)
{ // If both players have the same
// move then it's a draw
if (first == second)
return 0;
if (first == 'R' )
{
if (second == 'S' )
return 1;
else
return -1;
}
if (first == 'P' )
{
if (second == 'R' )
return 1;
else
return -1;
}
if (first == 'S' )
{
if (second == 'P' )
return 1;
else
return -1;
}
return 0;
} // Function that returns the count of games // won by both the players static Tuple< int , int > countWins( int k, string a,
string b)
{ int n = a.Length;
int m = b.Length;
int i = 0, j = 0;
// Total distinct games that
// can be played
int moves = n * m;
Tuple< int , int > wins = Tuple.Create(0, 0);
while (moves-- > 0)
{
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins = Tuple.Create(wins.Item1 + 1,
wins.Item2);
// Player 2 wins the current game
if (res == -1)
wins = Tuple.Create(wins.Item1,
wins.Item2 + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above
// n * m games repeat
int repeat = k / (n * m);
// Update the games won
wins = Tuple.Create(wins.Item1 * repeat,
wins.Item2 * repeat);
// Remaining number of games after
// removing repeated games
int rem = k % (n * m);
while (rem-- > 0)
{
int res = compare(a[i], b[j]);
// Player 1 wins the current game
if (res == 1)
wins = Tuple.Create(wins.Item1 + 1,
wins.Item2);
// Player 2 wins the current game
if (res == -1)
wins = Tuple.Create(wins.Item1,
wins.Item2 + 1);
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
} // Driver Code static void Main()
{ int k = 4;
string a = "SR" , b = "R" ;
Tuple< int , int > wins = countWins(k, a, b);
Console.WriteLine(wins.Item1 + " " +
wins.Item2);
} } // This code is contributed by divyesh072019 |
<script> // Javascript implementation of the above approach // Function that returns 1 if first player wins, // 0 in case of a draw and -1 if second player wins function compare(first,second)
{ // If both players have the same move
// then it's a draw
if (first == second)
return 0;
if (first == 'R ')
{
if (second == ' S ')
return 1;
else
return -1;
}
if (first == ' P ')
{
if (second == ' R ')
return 1;
else
return -1;
}
if (first == ' S ')
{
if (second == ' P')
return 1;
else
return -1;
}
return 0;
} // Function that returns the count of games // won by both the players function countWins(k,a,b)
{ let n = a.length;
let m = b.length;
let i = 0, j = 0;
// Total distinct games that
// can be played
let moves = n * m;
let wins = [0, 0];
while (moves-- > 0)
{
let res = compare(a[i],
b[j]);
// Player 1 wins the current game
if (res == 1)
wins = [wins[0] + 1,
wins[1]];
// Player 2 wins the current game
if (res == -1)
wins =[wins[0],
wins[1] + 1];
i = (i + 1) % n;
j = (j + 1) % m;
}
// Number of times the above
// n * m games repeat
let repeat = k / (n * m);
// Update the games won
wins = [wins[0] * repeat,
wins[1] * repeat];
// Remaining number of games after
// removing repeated games
let rem = k % (n * m);
while (rem-- > 0)
{
let res = compare(a[i],
b[j]);
// Player 1 wins the current game
if (res == 1)
wins = [wins[0] + 1,
wins[1]];
// Player 2 wins the current game
if (res == -1)
wins = [wins[0],
wins[1] + 1];
i = (i + 1) % n;
j = (j + 1) % m;
}
return wins;
} // Driver code let k = 4; let a = "SR" , b = "R" ;
let wins = countWins(k, a, b); document.write(wins[0] + " " + wins[1]);
// This code is contributed by avanitrachhadiya2155 </script> |
0 2
Time Complexity: O(N * M)
Auxiliary Space: O(1)