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Optimal Strategy for a Game | Special Gold Coin

Given a row of silver coins among which a special gold coin is present. Two players play the game, and with each move, a player has to remove a coin from either the left or the right end of the row and the player who removes the special coin wins the game. The task is to find the winner of the game.

Examples: 

Input: str = “GSSS” 
Output: First 
The first player directly removes the special gold coin.

Input: str = “SGS” 
Output: Second 
Irrespective of which coin the first player removes, the special 
gold coin becomes exposed and is removed by the second player. 
 

Approach: It can be observed by taking a few examples that if the count of the silver coins is odd then the first player wins the game otherwise the player two wins the game. In special case when Gold coin is in corner, First player will be the winner regardless the count of silver coins.

Below are the steps to implement the above approach:

Below is the implementation of the above approach: 




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// winner of the game
string getWinner(string str, int len)
{
     
    // To store the count of silver coins
    int total = 0;
     
    if(str[0]=='G' ||str[len-1]=='G')
        return "First";
    else{
        for (int i = 0; i < len; i++) {
 
            // Update the position of
           // the gold coin
            if (str[i] == 'S') {
                total++;
            }
        }
 
        // First player will win the game
        if ((total % 2) == 1)
            return "First";
        return "Second";
    }
}
 
// Driver code
int main()
{
    string str = "GSSS";
    int len = str.length();
 
    cout << getWinner(str, len);
 
    return 0;
}




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the
// winner of the game
static String getWinner(String str, int len)
{
 
    // To store the count of silver coins
    int total = 0;
    for (int i = 0; i < len; i++)
    {
 
        // Update the position of
        // the gold coin
        if (str.charAt(i) == 'S')
        {
            total++;
        }
    }
 
    // First player will win the game
    if ((total % 2) == 1)
        return "First";
    return "Second";
}
 
// Driver code
public static void main(String []args)
{
    String str = "GSSS";
    int len = str.length();
 
    System.out.println(getWinner(str, len));
}
}
 
// This code is contributed by Surendra_Gangwar




# Python3 implementation of the approach
 
# Function to return the
# winner of the game
def getWinner(string, length) :
 
    # To store the count of silver coins
    total = 0;
    for i in range(length) :
 
        # Update the position of
        # the gold coin
        if (string[i] == 'S') :
            total += 1;
 
    # First player will win the game
    if ((total % 2) == 1) :
        return "First";
    return "Second";
 
# Driver code
if __name__ == "__main__" :
 
    string = "GSSS";
    length = len(string);
 
    print(getWinner(string, length));
     
# This code is contributed by kanugargng




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the
// winner of the game
static String getWinner(String str, int len)
{
 
    // To store the count of silver coins
    int total = 0;
    for (int i = 0; i < len; i++)
    {
 
        // Update the position of
        // the gold coin
        if (str[i] == 'S')
        {
            total++;
        }
    }
 
    // First player will win the game
    if ((total % 2) == 1)
        return "First";
    return "Second";
}
 
// Driver code
public static void Main(string []args)
{
    string str = "GSSS";
    int len = str.Length;
 
    Console.WriteLine(getWinner(str, len));
}
}
 
// This code is contributed by rrrtnx.




<script>
 
// Javascript implementation of the approach
 
// Function to return the
// winner of the game
function getWinner(str, len)
{
     
    // To store the count of silver coins
    var total = 0;
     
    if (str[0] == 'G' || str[len - 1] == 'G')
        return "First";
    else
    {
        for(var i = 0; i < len; i++)
        {
            // Update the position of
            // the gold coin
            if (str[i] == 'S')
            {
                total++;
            }
        }
 
        // First player will win the game
        if ((total % 2) == 1)
            return "First";
             
        return "Second";
    }
}
 
// Driver code
var str = "GSSS";
var len = str.length;
 
document.write(getWinner(str, len));
 
// This code is contributed by importantly
 
</script>

Output: 
First

 

Time Complexity: O(n) where n is the size of the string.

Auxiliary Space: O(1)


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