README.md

794. Valid Tic-Tac-Toe State

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 ' '.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn valid_tic_tac_toe(board: Vec<String>) -> bool {
        let board = board.concat().into_bytes();
        let x_count = board.iter().filter(|&&c| c == b'X').count();
        let o_count = board.iter().filter(|&&c| c == b'O').count();

        (x_count == o_count + 1 && !Self::is_win(b'O', &board))
            || (x_count == o_count && !Self::is_win(b'X', &board))
    }

    fn is_win(player: u8, board: &[u8]) -> bool {
        [
            (0, 1, 2),
            (3, 4, 5),
            (6, 7, 8),
            (0, 3, 6),
            (1, 4, 7),
            (2, 5, 8),
            (0, 4, 8),
            (2, 4, 6),
        ]
        .iter()
        .any(|&(x, y, z)| [board[x], board[y], board[z]] == [player, player, player])
    }
}