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

implSolution{pubfnvalid_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))}fnis_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])}}