Two players are playing a game where string str is given. The first player can take the characters at even indices and the second player can take the characters at odd indices. The player which can build the lexicographically smaller string than the other player wins the game. Print the winner of the game, either player A, B or print Tie if it’s a tie.
Examples:
Input: str = “geeksforgeeks”
Output: B
Explanation: “eeggoss” is the lexicographically smallest
string that player A can get.
“eefkkr” is the lexicographically smallest
string that player B can get.
And B’s string is lexicographically smaller.Input: str = “abcdbh”
Output: A
Approach: Create two empty strings str1 and str2 for players A and B respectively. Traverse the original string character by character and for every character whose index is even, append this character in str1 else append this character in str2. Finally, sort the generated string in order to get the lexicographically smallest possible string and compare them to find the winner of the game.
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
B
Time Complexity: O(n log(n)), Where n is the length of the given string.
Auxiliary Space: O(n), for storing the strings for both the players.