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

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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

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>

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
?>

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.

Using a set:

Approach:

In this approach, we will first create a set of all the characters in the string. Then, we will count the number of characters with odd frequency. If this count is greater than 1, Player-2 will win, otherwise, Player-1 will win.

Create a set of all the unique characters in the string s.
For each character c in the set, count the number of occurrences of c in s.
If the count of any character is odd, increment the odd_freq_count variable.
If odd_freq_count is greater than 1, return “Player-2”, indicating that Player-2 will win the game.
Otherwise, return “Player-1”, indicating that Player-1 will win the game.

C++

#include <bits/stdc++.h>
#include <string>
#include <unordered_set>
 
std::string palindrome_game(std::string s) {
    std::unordered_set<char> chars;
 
    for (char c : s) {
        chars.insert(c);
    }
 
    int odd_freq_count = 0;
    for (char c : chars) {
        if (std::count(s.begin(), s.end(), c) % 2 != 0) {
            odd_freq_count++;
        }
    }
 
    if (odd_freq_count > 1) {
        return "Player-2";
    } else {
        return "Player-1";
    }
}
 
int main() {
    std::string s = "abaaab";
    std::string result = palindrome_game(s);
    std::cout << result << std::endl;  
 
    return 0;
}

Java

import java.util.HashMap;
import java.util.HashSet;
 
public class Main {
    public static String palindromeGame(String s) {
        // Create a HashSet to store unique characters in the string.
        HashSet<Character> chars = new HashSet<>();
 
        // Iterate through the string and add each character to the HashSet.
        for (char c : s.toCharArray()) {
            chars.add(c);
        }
 
        int oddFreqCount = 0;
        for (char c : chars) {
            // Count the frequency of each character in the string.
            int charCount = 0;
            for (char ch : s.toCharArray()) {
                if (ch == c) {
                    charCount++;
                }
            }
 
            // Check if the frequency is odd.
            if (charCount % 2 != 0) {
                oddFreqCount++;
            }
        }
 
        if (oddFreqCount > 1) {
            return "Player-2";
        } else {
            return "Player-1";
        }
    }
 
    public static void main(String[] args) {
        String s = "abaaab";
        String result = palindromeGame(s);
        System.out.println(result);
    }
}
 
// This code is contributed by akshitaguprzj3

Python3

def palindrome_game(s):
    chars = set(s)
 
    odd_freq_count = sum(1 for c in chars if s.count(c) % 2 != 0)
 
    if odd_freq_count > 1:
        return "Player-2"
    else:
        return "Player-1"
 
s = "abaaab"
result = palindrome_game(s)
print(result)  # Output: Player-1

C#

using System;
using System.Collections.Generic;
 
class Program {
    // Function to determine the winner of the palindrome
    // game
    static string PalindromeGame(string s)
    {
        // Create a HashSet to store unique characters
        HashSet<char> chars = new HashSet<char>();
 
        // Iterate through the string and insert characters
        // into the HashSet
        foreach(char c in s) { chars.Add(c); }
 
        int oddFreqCount = 0;
 
        // Iterate through the unique characters in the
        // HashSet
        foreach(char c in chars)
        {
            // Count the frequency of the character in the
            // original string
            int charFrequency = 0;
            foreach(char originalChar in s)
            {
                if (originalChar == c) {
                    charFrequency++;
                }
            }
 
            // Check if the character's frequency is odd
            if (charFrequency % 2 != 0) {
                oddFreqCount++;
            }
        }
 
        // Determine the winner based on odd frequency count
        if (oddFreqCount > 1) {
            return "Player-2";
        }
        else {
            return "Player-1";
        }
    }
 
    static void Main()
    {
        string s = "abaaab";
        string result = PalindromeGame(s);
        Console.WriteLine(result);
 
        Console.ReadKey();
    }
}

Javascript

function palindrome_game(s) {
    // Create a set of characters in the string
    let chars = new Set(s);
 
    // Count the number of characters with odd frequency
    let odd_freq_count = 0;
    for (let c of chars) {
        if ((s.split(c).length - 1) % 2 != 0) {
            odd_freq_count++;
        }
    }
 
    // If there are more than one character with odd frequency, Player-2 wins
    if (odd_freq_count > 1) {
        return "Player-2";
    } else {
        return "Player-1";
    }
}
 
let s = "abaaab";
let result = palindrome_game(s);
console.log(result);

Output

Player-1

Time Complexity: O(n), where n is the length of the string
Auxiliary Space: O(k), where k is the number of unique characters in the string

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