LeetCode - Design Tic-Tac-Toe - Day15 (1/15/2020)

Design Tic-Tac-Toe

Question

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?

What I though

Because I have solved the similar question of this, I set board[][] and try to manage it by adding 'X' and 'O' in move function. However, once we try to check who wins, it takes O(n²) run time to find the answer.

Solution

The easiest way to manage the current situation of Tic-Tac-Toe is to use Integer. Let's say if player 1 move his X, we increment the integer for the arrays matching with the coordinate. Instead, if player 2 move his O, we decrement.

In this condition, if one of the element in the array reach n, it means either player 1 or 2 win. In addition, we need to manage diagonal/anti-diagonal cases by using the same way.

class TicTacToe {
    private int[] rows;
    private int[] cols;
    private diagonal;
    private antiDiagonal;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[n];
    }
    
    /** 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) {
        int toAdd = player == 1 ? 1 : -1;
        
        rows[row] += toAdd;
        cols[col] += toAdd;
        if (row == col) {
            diagonal += toAdd;
        }

        if (col == (cols.length - row - 1)) {
            antiDiagonal += toAdd;
        }

        int size = rows.length;
        if (Math.abs(rows[row]) == size ||
            Math.abs(cols[col]) == size ||
            Math.abs(diagonal) == size  ||
            Math.abs(antiDiagonal) == size)
        {
            return player;
        }

        return 0;
    }
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */