Find the last player to be able to flip a character in a Binary String

Welcome to our comprehensive guide on a stimulating algorithmic problem involving binary strings. We will be examining a problem where we need to find the last player who can flip a character in a binary string. This problem is a great exercise for understanding game theory and binary string manipulation.

Problem Statement

Given a binary string, we have two players who take turns to flip a '1' into a '0'. The player who cannot make a move loses the game. The task is to find out who, between Player 1 and Player 2, will be the last player to be able to flip a character.

Approach

We will iterate over the binary string, counting the number of '1's. If the number of '1's is even, Player 2 will be the last to flip a '1', as Player 1 always starts the game. If the number of '1's is odd, Player 1 will be the last to flip a '1'.

Implementation

Example

Here're the programs to the above algorithm

 #include <stdio.h> #include <string.h> // Function to determine the last player to flip a character char* lastPlayer(char* s) { int count = 0; for (int i = 0; i < strlen(s); i++) { if (s[i] == '1') count++; } return (count % 2 == 0) ? "Player 2" : "Player 1"; } int main() { char s[] = "1101"; printf("The last player to be able to flip a character is: %s\n", lastPlayer(s)); return 0; } 

 The last player to be able to flip a character is: Player 1 

 #include<bits/stdc++.h> using namespace std; string lastPlayer(string s) { int count = 0; for (char c : s) { if (c == '1') count++; } return (count % 2 == 0) ? "Player 2" : "Player 1"; } int main() { string s="1101"; cout << "The last player to be able to flip a character is: " << lastPlayer(s) << endl; return 0; } 

 The last player to be able to flip a character is: Player 1 

 public class Main { // Function to determine the last player to flip a character public static String lastPlayer(String s) { int count = 0; for (char c : s.toCharArray()) { if (c == '1') count++; } return (count % 2 == 0) ? "Player 2" : "Player 1"; } public static void main(String[] args) { String s = "1101"; System.out.println("The last player to be able to flip a character is: " + lastPlayer(s)); } } 

 The last player to be able to flip a character is: Player 1 

 def last_player(s): count = s.count('1') return "Player 2" if count % 2 == 0 else "Player 1" def main(): s = "1101" print("The last player to be able to flip a character is:", last_player(s)) if __name__ == "__main__": main() 

 The last player to be able to flip a character is: Player 1 

This program inputs a binary string and outputs the last player who can flip a character.

Test Case Example

Let's consider an example to clarify this problem and its solution ?

Suppose the binary string is "1101".

• We start by counting the number of '1's in the binary string.

• The count of '1's in "1101" is 3, which is an odd number.

• Since the count is odd, Player 1 will be the last to flip a '1'.

• Therefore, the output will be "The last player to be able to flip a character is: Player 1".

Conclusion

In this guide, we've learned how to determine the last player who can flip a character in a binary string. This problem is an interesting exploration of game theory and binary string manipulation.