Question

You are going to write a method (to be called validateBoard) that is going to validate whether or not a Tic-Tac-Toe board is possible. Tic- Tac-Toe is played on a 3 x 3 board and players take turns placing either an x or an o on the board. We will assume that in Tic-Tac-Toe the player placing x will go first arld that o will go second. (Learn more about the game here) As the player placing x pieces goes first, and play alternates, a valid board is one in which there is either the same number or one more x game pieces on the board as there are o game pieces The board is represented by a 3 x 3 array that holds strings that represent the game piece at each location. A space can hold any of the following: name piece in that position

Answer 1

//Java program

import java.util.*;
public class PoD {
   public static void main(String args[]) {
       Scanner in = new Scanner(System.in);
       int gamepeice;
       String [][] ticTacBoard = new String[3][3];
       for(int i=0;i<3;i++) {
           for(int j=0;j<3;j++) {
              
               ticTacBoard[i][j] = in.next();
          
           if(i<2) {
               System.out.print(ticTacBoard[i][j]+" ");
           }
           else {
               System.out.print(ticTacBoard[i][j]+" ");
           }
       }
       }
       if(validateBoard(ticTacBoard)) {
           System.out.println("valid board");
       }
       else System.out.println("Invalid board");
      
       System.out.println("END OF OUTPUT");
       in.close();
       }

   private static boolean validateBoard(String[][] ticTacBoard) {
       int countx=0;
       int counto = 0;
       for(int i=0;i<3;i++) {
           for(int j=0;j<3;j++) {
               if(ticTacBoard[i][j]=="X"||ticTacBoard[i][j]=="x")countx++;
               if(ticTacBoard[i][j]=="O"||ticTacBoard[i][j]=="o")counto++;
           }
       }
       return (countx==counto ||(countx == counto+1));
   }
  
}