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

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Given a string consisting of lower case alphabets.
Rules of the Game: 
 

  • A player can choose a pair of similar consecutive characters and erase them.
  • There are two players playing the game, the player who makes the last move wins.

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. 
 

  • Each time we encounter a character that is different from the one present in the top of the stack we add it to the stack.
  • If the stack top and the next character match we pop the character from the stack and increment the count.
  • At the end, we just need to see who wins by checking count%2.

Below is the implementation of the above approach: 
 

C++




#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;
}


Java




// 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


Python3




# 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


C#




// 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


Javascript




<script>
    // Javascript implementation for above approach
     
    // Function to play the game
    // and find the winner
    function findWinner(s)
    {
        let i, count = 0, n;
        n = s.length;
        let 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.length == 0 ||
                st[st.length - 1] != s[i])
            {
                st.push(s[i]);
            }
            else
            {
                count++;
                st.pop();
            }
        }
 
        // Check who has won
        if (count % 2 == 0)
        {
            document.write("B");
        }
        else
        {
            document.write("A");
        }
    }
     
    let s = "kaak";
   
    findWinner(s);
 
// This code is contributed by divyesh072019.
</script>


Output: 

B

 

Time Complexity: O(n)

Auxiliary Space: O(n)



Last Updated : 03 Aug, 2022
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