Find the player who rearranges the characters to get a palindrome string first

Given an even length string S consisting of lower-case English alphabets only, we have two players playing the game. The rules are as follows:

  • the player wins the game, if, at any move, a player can re-arrange the characters of the string to get a palindrome string.
  • if the player cannot win the game, he has to remove any character from the string.

Both players play the game optimally with player-1 starting the game. The task is to print the winner of the game.

Examples:

Input: S = “abaaab”
Output: Player-1
Player-1 in the first step arranges the characters to get “aabbaa” and wins the game.

Input: S = “abca”
Output: Player-2
As the game is being played optimally, player-1 removes ‘a’ to get string “bca” which cannot be rearranged by player-2 in the second move to win the game.
Player-2 optimally removes ‘b’ and the string is now “ca”.
In the third move, player-1 removes “a” as he cannot rearrange the characters, the new string is “c”, which the player-2 at the next move can make a palindrome.



Approach:

  • Count the frequencies of each character in a freq[] array.
  • Count the characters that occur odd number of times.
  • If the count is 0 or an odd number, then Player-1 will always win the game, else player-2 will win because player-2 will make the last move.

Below is the implementation of the above approach:

C++

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// C++ program to print the winner of the game
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the winner of the game
int returnWinner(string s, int l)
{
    // Initialize the freq array to 0
    int freq[26];
    memset(freq, 0, sizeof freq);
  
    // Iterate and count the frequencies
    // of each character in the string
    for (int i = 0; i < l; i++) {
        freq[s[i] - 'a']++;
    }
  
    int cnt = 0;
  
    // Count the odd occurring character
    for (int i = 0; i < 26; i++) {
  
        // If odd occurrence
        if (freq[i] & 1)
            cnt++;
    }
  
    // Check condition for Player-1 winning the game
    if (cnt == 0 || cnt & 1)
        return 1;
    else
        return 2;
}
  
// Driver code
int main()
{
    string s = "abaaab";
    int l = s.length();
  
    // Function call that returns the winner
    int winner = returnWinner(s, l);
  
    cout << "Player-" << winner;
    return 0;
}

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Java

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// Java program to print the winner of the game 
class GfG {
// Function that returns the winner of the game 
static int returnWinner(String s, int l) 
    // Initialize the freq array to 0 
    int freq[]  =new int[26]; 
  
    // Iterate and count the frequencies 
    // of each character in the string 
    for (int i = 0; i < l; i++) { 
        freq[s.charAt(i) - 'a']++; 
    
  
    int cnt = 0
  
    // Count the odd occurring character 
    for (int i = 0; i < 26; i++) { 
  
        // If odd occurrence 
        if (freq[i] % 2 != 0
            cnt++; 
    
  
    // Check condition for Player-1 winning the game 
    if ((cnt == 0)|| (cnt & 1) == 1
        return 1
    else
        return 2
  
// Driver code 
public static void main(String[] args) 
    String s = "abaaab"
    int l = s.length(); 
  
    // Function call that returns the winner 
    int winner = returnWinner(s, l); 
  
    System.out.println("Player-" + winner); 
}

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PHP

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<?php
// PHP program to print the winner
// of the game 
  
// Function that returns the 
// winner of the game 
function returnWinner($s, $l
    // Initialize the freq array to 0 
    // int freq[26]; 
    // memset(freq, 0, sizeof freq); 
    // $freg=array_fill()
    $freq = array_fill(0, 26, 0);
      
    // Iterate and count the frequencies 
    // of each character in the string 
    for ($i = 0; $i < $l; $i++) 
    
        $freq[$s[$i] - 'a']++; 
    
  
    $cnt = 0; 
  
    // Count the odd occurring character 
    for ($i = 0; $i < 26; $i++) 
    
  
        // If odd occurrence 
        if ($freq[$i] & 1) 
            $cnt++; 
    
  
    // Check condition for Player-1 
    // winning the game 
    if ($cnt == 0 || $cnt & 1) 
        return 1; 
    else
        return 2; 
  
// Driver code 
$s = "abaaab"
$l = strlen($s); 
  
// Function call that returns 
// the winner 
$winner = returnWinner($s, $l); 
  
echo "Player-" , $winner
  
// This code is contributed
// by tushil
?>

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Output:

Player-1


My Personal Notes arrow_drop_up

Striver(underscore)79 at Codechef and codeforces D

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Improved By : jit_t, prerna saini