Leetcode 794. Valid Tic-Tac-Toe State

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O".  The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares (" ").
• The first player always places "X" characters, while the second player always places "O" characters.
• "X" and "O" characters are always placed into empty squares, never filled ones.
• The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
• The game also ends if all squares are non-empty.
• No more moves can be played if the game is over.

Example 1:
Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:
Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:
Input: board = ["XOX", "O O", "XOX"]
Output: true

Note:

• board is a length-3 array of strings, where each string board[i] has length 3.
• Each board[i][j] is a character in the set {" ", "X", "O"}
• 
  

Analysis:

The problem only asks if the given board is valid. A board is valid only if 

1. number of X - number of O == 0

2. number of X - number of O == 1

If the number of X is equal to number of O, we need to check player X does not win. Otherwise,  player O must be less than player X.

If the number of X - number of O = 1, we need to check player O does not win. Otherwise, the game has stopped and player X cannot be greater than player O.

Code (Java):

class Solution {
    public boolean validTicTacToe(String[] board) {
        char[][] gameBoard = new char[3][3];
        
        for (int i = 0; i < 3; i++) {
            gameBoard[i] = board[i].toCharArray();
        }
        
        // Get number of X player and O player
        //
        int numOfX = 0;
        int numOfO = 0;
        
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                char c = gameBoard[i][j];
                
                if (c == 'X') {
                    numOfX++;
                } else if (c == 'O') {
                    numOfO++;
                }
            }
        }
        
        if (numOfX < numOfO || numOfX - numOfO > 1) {
            return false;
        }
        
        if (numOfX - numOfO == 1) {
            if (hasWon(gameBoard, 'O')) {
                return false;
            }
        }
        
        if (numOfX == numOfO) {
            if (hasWon(gameBoard, 'X')) {
                return false;
            }
        }
        
        return true;
    }
    
    private boolean hasWon(char[][] gameBoard, char player) {
        for (int i = 0; i < 3; i++) {
            if (gameBoard[i][0] == player && 
                gameBoard[i][1] == gameBoard[i][0] && 
                gameBoard[i][2] == gameBoard[i][1]) {
                return true;
            }
            
            if (gameBoard[0][i] == player && 
                gameBoard[1][i] == gameBoard[0][i] && 
                gameBoard[2][i] == gameBoard[1][i]) {
                return true;
            }
            
            if (player == gameBoard[0][0] && 
                gameBoard[0][0] == gameBoard[1][1] && 
                gameBoard[1][1] == gameBoard[2][2]) {
                return true;
            }
            
            if (player == gameBoard[0][2] && 
                gameBoard[1][1] == gameBoard[0][2] && 
                gameBoard[1][1] == gameBoard[2][0]) {
                return true;
            }
        }
        
        return false;
    }
}