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Number of wins for each player in a series of Rock-Paper-Scissor game
  • Difficulty Level : Basic
  • Last Updated : 28 May, 2021

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 = Draw

Input: 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++




// 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
0 2

 

Time Complexity: O(N * M)
 

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