Design Tic-Tac-Toe Python Solution with Tests

lcpy gen -n 348   # by problem number
lcpy gen -s design_tic_tac_toe   # by problem name

Input
["TicTacToe", "move", "move", "move", "move", "move", "move", "move"]
[[3], [0, 0, 1], [0, 2, 2], [2, 2, 1], [1, 1, 2], [2, 0, 1], [1, 0, 2], [2, 1, 1]]
Output
[null, 0, 0, 0, 0, 0, 0, 1]

Explanation
TicTacToe ticTacToe = new TicTacToe(3);
Assume that player 1 is "X" and player 2 is "O" in the board.
ticTacToe.move(0, 0, 1); // return 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

ticTacToe.move(0, 2, 2); // return 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

ticTacToe.move(2, 2, 1); // return 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

ticTacToe.move(1, 1, 2); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

ticTacToe.move(2, 0, 1); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

ticTacToe.move(1, 0, 2); // return 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

ticTacToe.move(2, 1, 1); // return 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

class TicTacToe:
    # Time: O(n) setup — per-player row, column, and diagonal counters
    # Space: O(n) — 2n rows + 2n cols + 2 diagonals per player index
    def __init__(self, n: int) -> None:
        self.n = n
        self.rows = [[0] * n for _ in range(3)]
        self.cols = [[0] * n for _ in range(3)]
        self.diag = [0] * 3
        self.anti_diag = [0] * 3

    # Time: O(1) — increment four counters, compare against n
    # Space: O(1)
    def move(self, row: int, col: int, player: int) -> int:
        self.rows[player][row] += 1
        self.cols[player][col] += 1
        if row == col:
            self.diag[player] += 1
        if row + col == self.n - 1:
            self.anti_diag[player] += 1
        if (
            self.rows[player][row] == self.n
            or self.cols[player][col] == self.n
            or self.diag[player] == self.n
            or self.anti_diag[player] == self.n
        ):
            return player
        return 0