Find the winner of the game

Two players are playing a game where a string str is given. The first player can take the characters at even indices and the second player can take the characters at odd indices. The player which can build the lexicographically smaller string than the other player wins the game. Print the winner of the game, either player A, B or print Tie if its a tie.

Examples:

Input: str = “geeksforgeeks”
Output: B
“eeggoss” is the lexicographically smallest
string that player A can get.
“eefkkr” is the lexicographically smallest
string that player B can get.
And B’s string is lexicographically smaller.



Input: str = “abcdbh”
Output: A

Approach: Create two empty strings str1 and str2 for player A and B respectively. Traverse the original string character by character and for every character whose index is even, append this character in str1 else append this character in str2. Finally sort the generated string in order to get the lexicographically smallest possible string and compare them to find the winner of the game.

Below is the implementation of the above appoach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the winner of the game
void find_winner(string str, int n)
{
  
    // To store the strings for both the players
    string str1 = "", str2 = "";
    for (int i = 0; i < n; i++) {
  
        // If the index is even
        if (i % 2 == 0) {
  
            // Append the current character
            // to player A's string
            str1 += str[i];
        }
  
        // If the index is odd
        else {
  
            // Append the current character
            // to player B's string
            str2 += str[i];
        }
    }
  
    // Sort both the strings to get
    // the lexicographically smallest
    // string possible
    sort(str1.begin(), str1.end());
    sort(str2.begin(), str2.end());
  
    // Copmpare both the strings to
    // find the winner of the game
    if (str1 < str2)
        cout << "A";
    else if (str2 < str1)
        cout << "B";
    else
        cout << "Tie";
}
  
// Driver code
int main()
{
    string str = "geeksforgeeks";
    int n = str.length();
  
    find_winner(str, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.Arrays;
  
class GFG 
{
  
// Function to find the winner of the game
static void find_winner(String str, int n)
{
  
    // To store the strings for both the players
    String str1 = "", str2 = "";
    for (int i = 0; i < n; i++)
    {
  
        // If the index is even
        if (i % 2 == 0
        {
  
            // Append the current character
            // to player A's string
            str1 += str.charAt(i);
        }
  
        // If the index is odd
        else
        {
  
            // Append the current character
            // to player B's string
            str2 += str.charAt(i);
        }
    }
  
    // Sort both the strings to get
    // the lexicographically smallest
    // string possible
    char a[] = str1.toCharArray();
    Arrays.sort(a);
    char b[] = str2.toCharArray();
    Arrays.sort(b);
    str1 = new String(a);
    str2 = new String(b);
  
    // Copmpare both the strings to
    // find the winner of the game
    if (str1.compareTo(str2) < 0)
        System.out.print("A");
    else if (str1.compareTo(str2) > 0)
        System.out.print("B");
    else
        System.out.print("Tie");
}
  
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.length();
  
    find_winner(str, n);
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation of the approach 
  
# Function to find the winner of the game 
def find_winner(string, n) :
  
    # To store the strings for both the players 
    string1= ""; string2 = ""; 
    for i in range(n) :
  
        # If the index is even 
        if (i % 2 == 0) :
  
            # Append the current character 
            # to player A's string 
            string1 += string[i]; 
          
        # If the index is odd 
        else
              
            # Append the current character 
            # to player B's string 
            string2 += string[i];
              
    # Sort both the strings to get
    # the lexicographically smallest
    # string possible
    string1 = "".join(sorted(string1))
    string2 = "".join(sorted(string2))
      
    # Copmpare both the strings to 
    # find the winner of the game 
    if (string1 < string2) :
        print("A", end = ""); 
          
    elif (string2 < string1) :
        print("B", end = "");
          
    else :
        print("Tie", end = "");
  
# Driver code 
if __name__ == "__main__"
  
    string = "geeksforgeeks"
    n = len(string); 
  
    find_winner(string, n); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to find the winner of the game
static void find_winner(String str, int n)
{
  
    // To store the strings for both the players
    String str1 = "", str2 = "";
    for (int i = 0; i < n; i++)
    {
  
        // If the index is even
        if (i % 2 == 0) 
        {
  
            // Append the current character
            // to player A's string
            str1 += str[i];
        }
  
        // If the index is odd
        else
        {
  
            // Append the current character
            // to player B's string
            str2 += str[i];
        }
    }
  
    // Sort both the strings to get
    // the lexicographically smallest
    // string possible
    char []a = str1.ToCharArray();
    Array.Sort(a);
    char []b = str2.ToCharArray();
    Array.Sort(b);
    str1 = new String(a);
    str2 = new String(b);
  
    // Copmpare both the strings to
    // find the winner of the game
    if (str1.CompareTo(str2) < 0)
        Console.Write("A");
    else if (str1.CompareTo(str2) > 0)
        Console.Write("B");
    else
        Console.Write("Tie");
}
  
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.Length;
  
    find_winner(str, n);
}
}
  
// This code is contributed by Rajput-Ji

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Output:

B


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Improved By : AnkitRai01, Rajput-Ji

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