Find the winner of the Game to Win by erasing any two consecutive similar alphabets

Given a string consisting of lower case alphabets.

Rules of the Game:

The task is to find the winner if A goes first and both play optimally.



Examples:

Input: str = "kaak" 
Output: B
Explanation:
    Initial String: "kaak"
    A's turn:
        removes: "aa"
        Remaining String: "kk"
    B's turn:
        removes: "kk"
        Remaining String: ""
    Since B was the last one to play
    B is the winner.

Input: str = "kk"
Output: A

Approach: We can use a stack to simplify the problem.

Below is the implementation of the above approach:

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#include <bits/stdc++.h>
using namespace std;
  
// Function to play the game 
// and find the winner
void findWinner(string s)
{
    int i, count = 0, n;
    n = s.length();
    stack<char> st;
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) {
        if (st.empty() || st.top() != s[i]) {
            st.push(s[i]);
        }
        else {
            count++;
            st.pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0) {
        cout << "B" << endl;
    }
    else {
        cout << "A" << endl;
    }
}
  
// Driver code
int main()
{
    string s = "kaak";
  
    findWinner(s);
  
    return 0;
}
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// Java implementation for above approach
import java.util.*;
  
class GFG 
{
  
// Function to play the game 
// and find the winner
static void findWinner(String s) 
{
    int i, count = 0, n;
    n = s.length();
    Stack<Character> st = new Stack<Character>();
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) 
    {
        if (st.isEmpty() || 
            st.peek() != s.charAt(i))
        {
            st.push(s.charAt(i));
        
        else
        {
            count++;
            st.pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0
    {
        System.out.println("B");
    
    else
    {
        System.out.println("A");
    }
}
  
// Driver code
public static void main(String[] args) 
{
    String s = "kaak";
  
    findWinner(s);
}
  
// This code is contributed by Rajput-Ji
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# Python3 implementation of the approach 
  
# Function to play the game 
# and find the winner 
def findWinner(s) : 
  
    count = 0
    n = len(s); 
    st = []; 
  
    # ckecking the top of the stack with 
    # the i th character of the string 
    # add it to the stack if they are different 
    # otherwise increment count 
    for i in range(n) :
        if (len(st) == 0 or st[-1] != s[i]) : 
            st.append(s[i]); 
              
        else
            count += 1
            st.pop(); 
  
    # Check who has won 
    if (count % 2 == 0) :
        print("B"); 
      
    else :
        print("A"); 
          
# Driver code 
if __name__ == "__main__"
  
    s = "kaak"
  
    findWinner(s);
  
# This code is contributed by AnkitRai01
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// C# implementation for above approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
// Function to play the game 
// and find the winner
static void findWinner(String s) 
{
    int i, count = 0, n;
    n = s.Length;
    Stack<char> st = new Stack<char>();
  
    // ckecking the top of the stack with
    // the i th character of the string
    // add it to the stack if they are different
    // otherwise increment count
    for (i = 0; i < n; i++) 
    {
        if (st.Count == 0 || 
            st.Peek() != s[i])
        {
            st.Push(s[i]);
        
        else
        {
            count++;
            st.Pop();
        }
    }
  
    // Check who has won
    if (count % 2 == 0) 
    {
        Console.WriteLine("B");
    
    else
    {
        Console.WriteLine("A");
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    String s = "kaak";
  
    findWinner(s);
}
  
// This code is contributed by 29AjayKumar
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Output:
B



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