Designing Tic-Tac-Toe in Constant Time

The design of a tic-tac-toe is one of the most popular interview questions. It can be done in constant time and linear space. Here is the problem: https://leetcode.com/problems/design-tic-tac-toe/

348. Design Tic-Tac-Toe

Medium

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

You may assume the following rules:

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 is allowed.
3. 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?

Solution: have a counter variable for each row, each col and for each diagonal (hence two for the diagonals). This is linear-space. Increment the corresponding counters for player one, and check whether the counters have reached the max. Decrease the corresponding counters for player two, and check whether the counters have reached -max. If none of the above has happened, return zero (draw). That's it. Code is below, and Merry Christmas, ho-ho-ho!!! ACC.

public class TicTacToe

{

    private int dimension = 0;

    private int[] rows = null;

    private int[] cols = null;

    private int diag1 = 0;

    private int diag2 = 0;

    /** Initialize your data structure here. */

    public TicTacToe(int n)

    {

        dimension = n;

        rows = new int[dimension];

        cols = new int[dimension];

    }

    /** Player {player} makes a move at ({row}, {col}).

        @param row The row of the board.

        @param col The column of the board.

        @param player The player, can be either 1 or 2.

        @return The current winning condition, can be either:

                0: No one wins.

                1: Player 1 wins.

                2: Player 2 wins. */

    public int Move(int row, int col, int player)

    {

        if (player == 1)

        {

            rows[row]++;

            if (rows[row] == dimension) return 1;

            cols[col]++;

            if (cols[col] == dimension) return 1;

            if (row == col)

            {

                diag1++;

                if (diag1 == dimension) return 1;

            }

            if (row + col == dimension - 1)

            {

                diag2++;

                if (diag2 == dimension) return 1;

            }

        }

        else

        {

            rows[row]--;

            if (rows[row] == -dimension) return 2;

            cols[col]--;

            if (cols[col] == -dimension) return 2;

            if (row == col)

            {

                diag1--;

                if (diag1 == -dimension) return 2;

            }

            if (row + col == dimension - 1)

            {

                diag2--;

                if (diag2 == -dimension) return 2;

            }

        }

        return 0; 

    }

}