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 winnervoid 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 codeint main(){ string s = "kaak"; findWinner(s); return 0;} |
Java
// Java implementation for above approachimport java.util.*;class GFG {// Function to play the game // and find the winnerstatic 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 codepublic 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 approachusing System;using System.Collections.Generic;class GFG {// Function to play the game // and find the winnerstatic 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 codepublic 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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