Find the last player who can flip the characters in a binary string
Welcome to our comprehensive guide to interesting algorithmic problems involving binary strings in C. We are going to look at a problem where we need to find the last player who is able to flip a character in a binary string. This question is very useful for understanding game theory and binary string operations.
Problem Statement
Given a binary string, we have two players taking turns flipping a '1' into a '0'. The player who cannot perform the flip loses the game. The task is to find out whether player 1 or player 2 can flip a character last.
method
We will iterate over the binary string and count the number of "1"s. If the number of "1"s is an even number, player 2 will be the last person to flip a "1" since player 1 always starts the game. If the number of "1"s is an odd number, player 1 will be the last person to flip a "1".
C implementation
Example
This is the C solution to the problem−
#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
This program takes in a binary string and outputs the last player who can flip the character.
Test case example
Let us consider an example to clarify this problem and its solution -
Assume the binary string is "1101".
We first count the number of '1's in the binary string.
The number of "1"s in "1101" is 3, which is an odd number.
Since the count is odd, Player 1 will be the last one to flip the '1'.
So the output will be "The last player able to flip roles was: Player 1".
in conclusion
In this C guide we learned how to determine the last player who can flip a character in a binary string. This question is an interesting exploration of game theory and binary string manipulation.
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