Design Tic-Tac-Toe

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

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

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | | // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | | // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | | // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| | // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| | // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| | // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| | // Player 1 makes a move at (2, 1).
|X|X|X|
Follow up:
Could you do better than O(n2) per move() operation?

Hint:

Could you trade extra space such that move() operation can be done in O(1)?
You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

Initialize a 1-d array of size n for rows, cols, diagonals and reverse diagonals. Every time player moves, corresponding row, col, diagonal and reverse diagonal increases. Otherwise those coordinates decreases. If certain element equals n, we have a winner.

public class TicTacToe {
    int[] rows;
    int[] cols;
    int[] diags;
    int[] revDiags;
    int n;

    public TicTacToe (int n) {
        this.n = n;
        rows = new int[n];
        cols = new int[n];
        diags = new int[n];
        revDiags = new int[n];
    }

    public int move(int row, int col, int player) {
        if (player == 1) {
            rows[row]++;
            cols[col]++;
            if (row == col) {
                diags[row]++;
            }
            if (row + col == n - 1) {
                revDiags[row]++;
            }
            if (rows[row] == n
                || cols[col] == n
                || diags[row] == n
                || revDiags[row] == n) {
                return player;
            }
        } else {
            rows[row]--;
            cols[col]--;
            if (row == col) {
                diags[row]--;
            }
            if (row + col == n - 1) {
                revDiags[row]--;
            }
            if (rows[row] == -n
                || cols[col] == -n
                || diags[row] == -n
                || revDiags[row] == -n) {
                return player;
            }
        }
        return 0;
    }
}