794. Valid Tic-Tac-Toe State

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LeetCode

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Problem

Given a Tic-Tac-Toe board 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 that 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 three 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

Constraints:

• board.length == 3
• board[i].length == 3
• board[i][j] is either ‘X’, ‘O’, or ‘ ‘.

Code

class Solution {
    public boolean validTicTacToe(String[] board) {
        int turns = 0;
        boolean xwin = false;
        boolean owin = false;
        int[] rows = new int[3];
        int[] cols = new int[3];
        int diag = 0;
        int antidiag = 0;

        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (board[i].charAt(j) == 'X') {
                    turns++;
                    rows[i]++;
                    cols[j]++;
                    if (i == j) diag++;
                    if (i + j == 2) antidiag++;
                } else if (board[i].charAt(j) == 'O') {
                    turns--;
                    rows[i]--;
                    cols[j]--;
                    if (i == j) diag--;
                    if (i + j == 2) antidiag--;
                }
            }
        }

        xwin = rows[0] == 3 || rows[1] == 3 || rows[2] == 3 ||
                cols[0] == 3 || cols[1] == 3 || cols[2] == 3 ||
                diag == 3 || antidiag == 3;
        owin = rows[0] == -3 || rows[1] == -3 || rows[2] == -3 ||
                cols[0] == -3 || cols[1] == -3 || cols[2] == -3 ||
                diag == -3 || antidiag == -3;

        if (xwin && turns == 0 || owin && turns == 1) {
            return false;
        }
        return (turns == 0 || turns == 1) && (!xwin || !owin);
    }
}