Dos jugadores están jugando un juego en el que se da una string str . El primer jugador puede tomar los caracteres en índices pares y el segundo jugador puede tomar los caracteres en índices impares. El jugador que puede construir la string lexicográficamente más pequeña que el otro jugador gana el juego. Imprime el ganador del juego, ya sea el jugador A , B o imprime Empate si es un empate.
Ejemplos:
Entrada: str = “geeksforgeeks”
Salida: B
“eeggoss” es la
string lexicográficamente más pequeña que puede obtener el jugador A.
“eefkkr” es la
string lexicográficamente más pequeña que puede obtener el jugador B.
Y la string de B es lexicográficamente más pequeña.
Entrada: str = “abcdbh”
Salida: A
Enfoque: Cree dos strings vacías str1 y str2 para el jugador A y B respectivamente. Recorra la string original carácter por carácter y para cada carácter cuyo índice sea par, agregue este carácter en str1 ; de lo contrario, agregue este carácter en str2 . Finalmente ordene la string generada para obtener la string lexicográficamente más pequeña posible y compárela para encontrar el ganador del juego.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the winner of the game
void find_winner(string str, int n)
{
// To store the strings for both the players
string str1 = "", str2 = "";
for (int i = 0; i < n; i++) {
// If the index is even
if (i % 2 == 0) {
// Append the current character
// to player A's string
str1 += str[i];
}
// If the index is odd
else {
// Append the current character
// to player B's string
str2 += str[i];
}
}
// Sort both the strings to get
// the lexicographically smallest
// string possible
sort(str1.begin(), str1.end());
sort(str2.begin(), str2.end());
// Compare both the strings to
// find the winner of the game
if (str1 < str2)
cout << "A";
else if (str2 < str1)
cout << "B";
else
cout << "Tie";
}
// Driver code
int main()
{
string str = "geeksforgeeks";
int n = str.length();
find_winner(str, n);
return 0;
}
Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Function to find the winner of the game
static void find_winner(String str, int n)
{
// To store the strings for both the players
String str1 = "", str2 = "";
for (int i = 0; i < n; i++)
{
// If the index is even
if (i % 2 == 0)
{
// Append the current character
// to player A's string
str1 += str.charAt(i);
}
// If the index is odd
else
{
// Append the current character
// to player B's string
str2 += str.charAt(i);
}
}
// Sort both the strings to get
// the lexicographically smallest
// string possible
char a[] = str1.toCharArray();
Arrays.sort(a);
char b[] = str2.toCharArray();
Arrays.sort(b);
str1 = new String(a);
str2 = new String(b);
// Compare both the strings to
// find the winner of the game
if (str1.compareTo(str2) < 0)
System.out.print("A");
else if (str1.compareTo(str2) > 0)
System.out.print("B");
else
System.out.print("Tie");
}
// Driver code
public static void main(String[] args)
{
String str = "geeksforgeeks";
int n = str.length();
find_winner(str, n);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach
# Function to find the winner of the game
def find_winner(string, n) :
# To store the strings for both the players
string1= ""; string2 = "";
for i in range(n) :
# If the index is even
if (i % 2 == 0) :
# Append the current character
# to player A's string
string1 += string[i];
# If the index is odd
else :
# Append the current character
# to player B's string
string2 += string[i];
# Sort both the strings to get
# the lexicographically smallest
# string possible
string1 = "".join(sorted(string1))
string2 = "".join(sorted(string2))
# Compare both the strings to
# find the winner of the game
if (string1 < string2) :
print("A", end = "");
elif (string2 < string1) :
print("B", end = "");
else :
print("Tie", end = "");
# Driver code
if __name__ == "__main__" :
string = "geeksforgeeks";
n = len(string);
find_winner(string, n);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to find the winner of the game
static void find_winner(String str, int n)
{
// To store the strings for both the players
String str1 = "", str2 = "";
for (int i = 0; i < n; i++)
{
// If the index is even
if (i % 2 == 0)
{
// Append the current character
// to player A's string
str1 += str[i];
}
// If the index is odd
else
{
// Append the current character
// to player B's string
str2 += str[i];
}
}
// Sort both the strings to get
// the lexicographically smallest
// string possible
char []a = str1.ToCharArray();
Array.Sort(a);
char []b = str2.ToCharArray();
Array.Sort(b);
str1 = new String(a);
str2 = new String(b);
// Compare both the strings to
// find the winner of the game
if (str1.CompareTo(str2) < 0)
Console.Write("A");
else if (str1.CompareTo(str2) > 0)
Console.Write("B");
else
Console.Write("Tie");
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
int n = str.Length;
find_winner(str, n);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript implementation of the approach
// Function to find the winner of the game
function find_winner(str, n)
{
// To store the strings for both the players
var str1 = "", str2 = "";
for (var i = 0; i < n; i++) {
// If the index is even
if (i % 2 == 0) {
// Append the current character
// to player A's string
str1 += str[i];
}
// If the index is odd
else {
// Append the current character
// to player B's string
str2 += str[i];
}
}
// Sort both the strings to get
// the lexicographically smallest
// string possible
str1 = str1.split('').sort();
str2 = str2.split('').sort();
// Compare both the strings to
// find the winner of the game
if (str1 < str2)
document.write( "A");
else if (str2 < str1)
document.write( "B");
else
document.write( "Tie");
}
// Driver code
var str = "geeksforgeeks";
var n = str.length;
find_winner(str, n);
</script>
Producción:
B