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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;
}
Output
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;
}
Output
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));
}
}
Output
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()
Output
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.
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