Winner in the RockPaperScissor game using Bit manipulation (original) (raw)

Winner in the Rock-Paper-Scissor game using Bit manipulation

Last Updated : 3 Aug, 2022

Two players are playing a series of games of Rock–paper–scissors. There are a total of N games played represented by an array arr[][] where arr[i][0] is the move of player one and arr[i][1] is the move of the player two in the ith game from the set {'R', 'P', 'S'}. The task is to find the winner of each of the game. Note that the game is a draw if both players choose the same item.
Examples:

Input: arr[] = {"RS", "SR", "SP", "PP"}
Output:
A
B
A
DRAW
Input: arr[] = {"SS", "RP", "PS"}
Output:
Draw
B
B

Approach: Suppose player one is represented by bit 1 and player two is represented by 0. Moreover, let Rock be represented by 00 (0 in decimal), Paper by 01 (1 in decimal) and Scissors with 10 (2 in decimal).
If player one chooses rock, it will be represented by 100,
Similarly, 101 means Paper is chosen by player one.
The first bit indicates the player number and the next two bits for their choice.
Pattern:
100 (4 in decimal) (player 1, rock), 001 (1 in decimal) (player 2, paper) -> player 2 won (4-1 = 3)
101 (5 in decimal) (player 1, paper), 010 (2 in decimal) (player 2, scissors) -> player 2 won (5-2 = 3)
110 (6 in decimal) (player 1, scissors), 000 (0 in decimal) (player 2, rock) -> player 2 won (6-0 = 6)
101 (5 in decimal) (player 1, paper), 000 (0 in decimal) (player 2, rock) -> player 1 won (5-0 = 5)
110 (6 in decimal) (player 1, scissors), 001 (1 in decimal) (player 2, paper) -> player 1 won (6-1 = 5)
100 (4 in decimal) (player 1, rock), 010 (2 in decimal) (player 2, scissors) -> player 1 won (4-2 = 2)
According to the pattern, if the difference is a multiple of 3 then player two wins or if the difference is 4 then the game is a draw. In the rest of the cases, player one wins the game.
Below is the implementation of the above approach:

C++ `

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;

// Function to return the // winner of the game string winner(string moves) { map<char, int> data; data['R'] = 0; data['P'] = 1; data['S'] = 2;

// Both the players chose to
// play the same move
if (moves[0] == moves[1]) {
    return "Draw";
}

// Player A wins the game
if (((data[moves[0]] | 1 << (2))
     - (data[moves[1]] | 0 << (2)))
    % 3) {
    return "A";
}

return "B";

}

// Function to perform the queries void performQueries(string arr[], int n) { for (int i = 0; i < n; i++) cout << winner(arr[i]) << endl; }

// Driver code int main() { string arr[] = { "RS", "SR", "SP", "PP" }; int n = sizeof(arr) / sizeof(string);

performQueries(arr, n);

return 0;

}

Java

// Java implementation of the approach import java.util.*;

class GFG {

// Function to return the // winner of the game static String winner(String moves) { HashMap<Character, Integer> data = new HashMap<Character, Integer>(); data.put('R', 0); data.put('P', 1); data.put('S', 2);

// Both the players chose to
// play the same move
if (moves.charAt(0) == moves.charAt(1))
{
    return "Draw";
}

// Player A wins the game
if (((data.get(moves.charAt(0)) | 1 << (2)) - 
     (data.get(moves.charAt(1)) | 0 << (2))) % 3 != 0) 
{
    return "A";
}

return "B";

}

// Function to perform the queries static void performQueries(String arr[], int n) { for (int i = 0; i < n; i++) System.out.print(winner(arr[i]) + "\n"); }

// Driver code public static void main(String[] args) { String arr[] = { "RS", "SR", "SP", "PP" }; int n = arr.length;

performQueries(arr, n);

} }

// This code is contributed by 29AjayKumar

Python3

Python3 implementation of the approach

Function to return the

winner of the game

def winner(moves): data = dict() data['R'] = 0 data['P'] = 1 data['S'] = 2

# Both the players chose to
# play the same move
if (moves[0] == moves[1]):
    return "Draw"

# Player A wins the game
if (((data[moves[0]] | 1 << (2)) - 
     (data[moves[1]] | 0 << (2))) % 3):
    return "A"

return "B"

Function to perform the queries

def performQueries(arr,n): for i in range(n): print(winner(arr[i]))

Driver code

arr = ["RS", "SR", "SP", "PP"] n = len(arr)

performQueries(arr, n)

This code is contributed by Mohit Kumar

C#

// C# implementation of the approach using System; using System.Collections.Generic;

class GFG {

// Function to return the // winner of the game static String winner(String moves) { Dictionary<char, int> data = new Dictionary<char, int>(); data.Add('R', 0); data.Add('P', 1); data.Add('S', 2);

// Both the players chose to
// play the same move
if (moves[0] == moves[1])
{
    return "Draw";
}

// Player A wins the game
if (((data[moves[0]] | 1 << (2)) - 
     (data[moves[1]] | 0 << (2))) % 3 != 0) 
{
    return "A";
}

return "B";

}

// Function to perform the queries static void performQueries(String []arr, int n) { for (int i = 0; i < n; i++) Console.Write(winner(arr[i]) + "\n"); }

// Driver code public static void Main(String[] args) { String []arr = { "RS", "SR", "SP", "PP" }; int n = arr.Length;

performQueries(arr, n);

} }

// This code is contributed by 29AjayKumar

JavaScript

`

Time Complexity: O(n)

Auxiliary Space: O(1)