Given a string S consisting of lower-case English alphabets only, we have two players playing the game. The rules are as follows:
- The player can remove any character from the given string S and write it on paper on any side(left or right) of an empty string.
- The player wins the game, if at any move he can get a palindromic string first of length > 1.
- If a palindromic string cannot be formed, Player-2 is declared the winner.
Both play optimally with player-1 starting the game. The task is to find the winner of the game.
Examples:
Input: S = “abc”
Output: Player-2
Explanation:
There are all unique characters due to which there
is no way to form a palindromic string of length > 1Input: S = “abccab”
Output: Player-2
Explanation:
Initially, newString = “” is empty.
Let Player-1 choose character ‘a’ and write it on paper.
Then, S = “bccab” and newString = “a”.
Now Player-2 chooses character ‘a’ from S and writes it on the left side of newString.
Thus, S = “bccb” and newString = “aa”.
Now, newString = “aa” is a palindrome of length 2.
Hence, Player-2 wins.
Approach: The idea is to formulate a condition in which Player-1 is always going to be the winner. If the condition fails, then Player-2 will win the game.
- If there is only one unique character occurring once in the given string, and the rest of the characters occurring more than 1, then Player-1 is going to be the winner, else Player-2 will always win.
- If we have all characters that are occurring more than once in the given string, then Player-2 can always copy the Player-1 move in his first turn and wins.
- Also, if we have more than one character in the string occurring one time only, then a palindrome string can never be formed(in the optimal case). Hence, again, Player-2 wins.
Below is the implementation of the above approach:
// C++ Implementation to find // which player can form a palindromic // string first in a game #include <bits/stdc++.h> using namespace std;
// Function to find // winner of the game int palindromeWinner(string& S)
{ // Array to Maintain frequency
// of the characters in S
int freq[26];
// Initialise freq array with 0
memset (freq, 0, sizeof freq);
// Maintain count of all
// distinct characters
int count = 0;
// Finding frequency of each character
for ( int i = 0; i < ( int )S.length();
++i) {
if (freq[S[i] - 'a' ] == 0)
count++;
freq[S[i] - 'a' ]++;
}
// Count unique duplicate
// characters
int unique = 0;
int duplicate = 0;
// Loop to count the unique
// duplicate characters
for ( int i = 0; i < 26; ++i) {
if (freq[i] == 1)
unique++;
else if (freq[i] >= 2)
duplicate++;
}
// Condition for Player-1
// to be winner
if (unique == 1 &&
(unique + duplicate) == count)
return 1;
// Else Player-2 is
// always winner
return 2;
} // Driven Code int main()
{ string S = "abcbc" ;
// Function call
cout << "Player-"
<< palindromeWinner(S)
<< endl;
return 0;
} |
// Java implementation to find which // player can form a palindromic // string first in a game import java.util.*;
class GFG{
// Function to find // winner of the game static int palindromeWinner(String S)
{ // Array to maintain frequency
// of the characters in S
int freq[] = new int [ 26 ];
// Initialise freq array with 0
Arrays.fill(freq, 0 );
// Maintain count of all
// distinct characters
int count = 0 ;
// Finding frequency of each character
for ( int i = 0 ; i < ( int )S.length(); ++i)
{
if (freq[S.charAt(i) - 'a' ] == 0 )
count++;
freq[S.charAt(i) - 'a' ]++;
}
// Count unique duplicate
// characters
int unique = 0 ;
int duplicate = 0 ;
// Loop to count the unique
// duplicate characters
for ( int i = 0 ; i < 26 ; ++i)
{
if (freq[i] == 1 )
unique++;
else if (freq[i] >= 2 )
duplicate++;
}
// Condition for Player-1
// to be winner
if (unique == 1 &&
(unique + duplicate) == count)
return 1 ;
// Else Player-2 is
// always winner
return 2 ;
} // Driver Code public static void main(String s[])
{ String S = "abcbc" ;
// Function call
System.out.println( "Player-" + palindromeWinner(S));
} } // This code is contributed by rutvik_56 |
# Python3 implementation to find # which player can form a palindromic # string first in a game # Function to find # winner of the game def palindromeWinner(S):
# Array to Maintain frequency
# of the characters in S
# initialise freq array with 0
freq = [ 0 for i in range ( 0 , 26 )]
# Maintain count of all
# distinct characters
count = 0
# Finding frequency of each character
for i in range ( 0 , len (S)):
if (freq[ ord (S[i]) - 97 ] = = 0 ):
count + = 1
freq[ ord (S[i]) - 97 ] + = 1
# Count unique duplicate
# characters
unique = 0
duplicate = 0
# Loop to count the unique
# duplicate characters
for i in range ( 0 , 26 ):
if (freq[i] = = 1 ):
unique + = 1
elif (freq[i] > = 2 ):
duplicate + = 1
# Condition for Player-1
# to be winner
if (unique = = 1 and
(unique + duplicate) = = count):
return 1
# Else Player-2 is
# always winner
return 2
# Driven Code S = "abcbc" ;
# Function call print ( "Player-" , palindromeWinner(S))
# This code is contributed by Sanjit_Prasad |
// C# implementation to find which // player can form a palindromic // string first in a game using System;
class GFG{
// Function to find // winner of the game static int palindromeWinner( string S)
{ // Array to maintain frequency
// of the characters in S
int [] freq = new int [26];
// Maintain count of all
// distinct characters
int count = 0;
// Finding frequency of
// each character
for ( int i = 0;
i < ( int )S.Length; ++i)
{
if (freq[S[i] - 'a' ] == 0)
count++;
freq[S[i] - 'a' ]++;
}
// Count unique duplicate
// characters
int unique = 0;
int duplicate = 0;
// Loop to count the unique
// duplicate characters
for ( int i = 0; i < 26; ++i)
{
if (freq[i] == 1)
unique++;
else if (freq[i] >= 2)
duplicate++;
}
// Condition for Player-1
// to be winner
if (unique == 1 &&
(unique + duplicate) ==
count)
return 1;
// Else Player-2 is
// always winner
return 2;
} // Driver Code public static void Main( string [] s)
{ string S = "abcbc" ;
// Function call
Console.Write( "Player-" +
palindromeWinner(S));
} } // This code is contributed by Chitranayal |
<script> // Javascript implementation to find which // player can form a palindromic // string first in a game // Function to find // winner of the game function palindromeWinner(S)
{ // Array to maintain frequency
// of the characters in S
let freq = new Array(26);
// Initialise freq array with 0
for (let i=0;i<26;i++)
{
freq[i]=0;
}
// Maintain count of all
// distinct characters
let count = 0;
// Finding frequency of each character
for (let i = 0; i < S.length; ++i)
{
if (freq[S[i].charCodeAt(0) - 'a' .charCodeAt(0)] == 0)
count++;
freq[S[i].charCodeAt(0) - 'a' .charCodeAt(0)]++;
}
// Count unique duplicate
// characters
let unique = 0;
let duplicate = 0;
// Loop to count the unique
// duplicate characters
for (let i = 0; i < 26; ++i)
{
if (freq[i] == 1)
unique++;
else if (freq[i] >= 2)
duplicate++;
}
// Condition for Player-1
// to be winner
if (unique == 1 &&
(unique + duplicate) == count)
return 1;
// Else Player-2 is
// always winner
return 2;
} // Driver Code let S = "abcbc" ;
// Function call document.write( "Player-" + palindromeWinner(S));
// This code is contributed by avanitrachhadiya2155 </script> |
Player-1
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.