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++

// 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;
}

Java

// 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);
}
}

Python3

# Python 3 program to print the winner of the game

# Function that returns the winner of the game
def returnWinner(s, l):

    # Initialize the freq array to 0
    freq = [0 for i in range(26)]

    # Iterate and count the frequencies
    # of each character in the string
    for i in range(0, l, 1):
        freq[ord(s[i]) - ord('a')] += 1

    cnt = 0

    # Count the odd occurring character
    for i in range(26):

        # If odd occurrence
        if (freq[i] % 2 != 0):
            cnt += 1

    # Check condition for Player-1
    # winning the game
    if (cnt == 0 or cnt & 1 == 1):
        return 1
    else:
        return 2

# Driver code
if __name__ == '__main__':
    s = "abaaab"
    l = len(s)

    # Function call that returns the winner
    winner = returnWinner(s, l)

    print("Player-", winner)

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to print the winner of the game
using System;

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[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);

    Console.WriteLine("Player-" + winner);
}
}

// This code contributed by Rajput-Ji

PHP

<?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
?>

Javascript

<script>
    // Javascript 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
        let freq = new Array(26);
        freq.fill(0);

        // Iterate and count the frequencies
        // of each character in the string
        for (let i = 0; i < l; i++)
        {
            freq[s[i].charCodeAt() - 'a'.charCodeAt()]++;
        }

        let cnt = 0;

        // Count the odd occurring character
        for (let 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;
    }

    let s = "abaaab";
    let l = s.length;

    // Function call that returns the winner
    let winner = returnWinner(s, l);

    document.write("Player-" + winner);

</script>

Output

Player-1

Complexity Analysis:

• Time Complexity: O(l), where l is the length of the string. As, we are using a loop to traverse l times.
• Auxiliary Space: O(26), as we are using extra space for storing the frequencies.