348. design tic tac toe

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves are allowed.
3. A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

• TicTacToe(int n) Initializes the object the size of the board n.
• int move(int row, int col, int player) Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move, and the two players alternate in making moves. Return
  • 0 if there is no winner after the move,
  • 1 if player 1 is the winner after the move, or
  • 2 if player 2 is the winner after the move.

solution

The logic for the optimization here is basically identical to the one used in 37-sudoku-solver.

class TicTacToe:
	def __init__(self, n: int):
		self.n = n
		self.rows = [[0,0] for _ in range(n)]
		self.cols = [[0,0] for _ in range(n)]
		# [ascending, descending]
		self.diags = [[0,0] for _ in range(2)]
	  
	def move(self, row: int, col: int, player: int) -> int:
		self.rows[row][player-1] += 1
		self.cols[col][player-1] += 1
	  
		if row == col:
			self.diags[1][player-1] += 1
		if row+col == self.n-1:
			self.diags[0][player-1] += 1
 
		if (
			self.rows[row][player-1] == self.n
			or self.cols[col][player-1] == self.n
			or self.diags[0][player-1] == self.n
			or self.diags[1][player-1] == self.n
		):
			return player
		return 0