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 gameint 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 codeint 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 gamedef 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 codeif __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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
