Given a numeric string str, the task is to determine the winner of the game when two players play a game optimally with the string as per the given conditions:
- Player 1 always starts first.
- In one turn, one player can remove the contiguous elements from the string and the number of elements removed will add up to his score. Player 1 will remove odd contiguous elements and Player 1 will remove even contiguous elements.
- Any player who is not able to make a move loses the game.
- After all the strings are removed, the player with the maximum score wins the game and if the scores are equal, then print “-1”.
Examples:
Input: str = “7788”
Output: -1
Explanation:
The first move for player 1 is to remove 77 and for player 2 it is 88. The score for both is 2. Hence -1.Input: str = “8822”
Output: Player 2
Explanation:
There are no odd elements, so player 1 can’t make a move, so player 2 wins.
Approach: Follow the steps below in order to solve the problem:
- Create an array A[] of size 10 to store the frequency of all the digits in the given string.
- Iterate the given string and update the frequency of each digit in the above array.
- Traverse the array A[] and take two variables as sum1 = 0 and sum2 = 0 to store the sum of scores.
- Increment sum1 if the index is odd i.e., Player 1 turn, else increment sum2 if the index is even i.e., Player 2 turn.
- After the above steps, check if sum1 is equal to sum2 then print -1 otherwise print the number of players who win the game on the basis of the score.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the winner of the// game when both players play optimallyvoid determineWinner(string str){ // Stores the frequency of all digit vector<int> A(10); // Stores the scores of player1 // and player2 respectively int sum1 = 0, sum2 = 0; // Iterate to store frequencies for (int i = 0; i < str.length(); i++) { A[int(str[i]) - 48]++; } for (int i = 0; i <= 9; i++) { // Turn for the player1 if (i % 2 != 0) { // Add score of player1 sum1 = sum1 + A[i]; } else { // Add score of player2 sum2 = sum2 + A[i]; } } // Check if its a draw if (sum1 == sum2) { cout << "-1"; } // If score of player 1 is greater else if (sum1 > sum2) { cout << "Player 1"; } // Otherwise else { cout << "Player 2"; }}// Driver Codeint main(){ string str = "78787"; determineWinner(str); return 0;} |
Java
// Java program for the above approach import java.util.*;class GFG{// Function to find the winner of the// game when both players play optimallystatic void determineWinner(String str){ // Stores the frequency of all digit int[] A = new int[10]; // Stores the scores of player1 // and player2 respectively int sum1 = 0, sum2 = 0; // Iterate to store frequencies for(int i = 0; i < str.length(); i++) { A[(int)str.charAt(i) - 48]++; } for(int i = 0; i <= 9; i++) { // Turn for the player1 if (i % 2 != 0) { // Add score of player1 sum1 = sum1 + A[i]; } else { // Add score of player2 sum2 = sum2 + A[i]; } } // Check if its a draw if (sum1 == sum2) { System.out.print("-1"); } // If score of player 1 is greater else if (sum1 > sum2) { System.out.print("Player 1"); } // Otherwise else { System.out.print("Player 2"); }}// Driver codepublic static void main (String[] args){ String str = "78787"; determineWinner(str);}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach # Function to find the winner of the# game when both players play optimallydef determineWinner(str): # Stores the frequency of all digit A = [0 for i in range(9)]; # Stores the scores of player1 # and player2 respectively sum1 = 0; sum2 = 0; # Iterate to store frequencies for i in range(0, len(str)): A[int(str[i])] += 1; for i in range(0, 9): # Turn for the player1 if (i % 2 != 0): # Add score of player1 sum1 = sum1 + A[i]; else: # Add score of player2 sum2 = sum2 + A[i]; # Check if its a draw if (sum1 == sum2): print("-1"); # If score of player 1 is greater elif(sum1 > sum2): print("Player 1"); # Otherwise else: print("Player 2"); # Driver codeif __name__ == '__main__': str = "78787"; determineWinner(str); # This code is contributed by Amit Katiyar |
C#
// C# program for the above approach using System;class GFG{// Function to find the winner of the// game when both players play optimallystatic void determineWinner(String str){ // Stores the frequency of all digit int[] A = new int[10]; // Stores the scores of player1 // and player2 respectively int sum1 = 0, sum2 = 0; // Iterate to store frequencies for(int i = 0; i < str.Length; i++) { A[(int)str[i] - 48]++; } for(int i = 0; i <= 9; i++) { // Turn for the player1 if (i % 2 != 0) { // Add score of player1 sum1 = sum1 + A[i]; } else { // Add score of player2 sum2 = sum2 + A[i]; } } // Check if its a draw if (sum1 == sum2) { Console.Write("-1"); } // If score of player 1 is greater else if (sum1 > sum2) { Console.Write("Player 1"); } // Otherwise else { Console.Write("Player 2"); }}// Driver codepublic static void Main(String[] args){ String str = "78787"; determineWinner(str);}}// This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the winner of the// game when both players play optimallyfunction determineWinner(str){ // Stores the frequency of all digit let A = Array.from({length: 10}, (_, i) => 0); // Stores the scores of player1 // and player2 respectively let sum1 = 0, sum2 = 0; // Iterate to store frequencies for(let i = 0; i < str.length; i++) { A[str[i].charCodeAt() - 48]++; } for(let i = 0; i <= 9; i++) { // Turn for the player1 if (i % 2 != 0) { // Add score of player1 sum1 = sum1 + A[i]; } else { // Add score of player2 sum2 = sum2 + A[i]; } } // Check if its a draw if (sum1 == sum2) { document.write("-1"); } // If score of player 1 is greater else if (sum1 > sum2) { document.write("Player 1"); } // Otherwise else { document.write("Player 2"); }}// Driver code let str = "78787"; determineWinner(str);</script> |
Output:
Player 1
Time Complexity: O(N)
Auxiliary Space: O(1) since it is using constant space
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