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)
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