348. Design Tic-Tac-Toe (Medium) Google Microsoft

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(n^2) permove()operation?

Solution 1: Brute Force O(n); O(n^2)

    int[][] board;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        board = new int[n][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) {
        board[row][col] = player; //use 1 to represents what palyer 1 places, use 2 for player2
        int i = 0, j = 0, n = board.length;
        for (; i < n; i++) { //check if current column is all filled with the mark which current palyer places.
            if (board[i][col] != player) break;
        }
        if (i == n) return player;

        for (; j < n; j++) {
            if (board[row][j] != player) break;
        }
        if (j == n) return player;

        if (row == col) {
            for (i = 0; i < n; i++) {
                if (board[i][i] != player) break;
            }
        }
        if (i == n) return player;

        if (row + col == n - 1) {
            for (i = 0; i < n; i++) {
                if (board[i][n - 1 - i] != player) break;
            }
        }
        if (i == n) return player;

        return 0;
    }

Solution 2: O(1); O(n)

Thus, we don’t need to keep track of an entire n^2 board. We only need to keep a count for each row and column. If at any time a row or column matches the size of the board then that player has won.

To keep track of which player, I add 1 for Player1 and -1 for Player2. There are two additional variables to keep track of the count of the diagonals. Each time a player places a piece we just need to check the count of that row, column, diagonal and anti-diagonal.

    /**
     * O(1); O(n)
     * Use 2 arrays to record each row and column and 2 variables to record diagonal and antidiagonal.
     * If a row or column or diagonal or antidiagonal matches the length of the board, current player has won.
     */
    private int[] rows;
    private int[] cols;
    private int diagonal;
    private int antiDiagonal;

    //    /** Initialize your data structure here. */
    public DesignTicTacToe(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;
        }
        int n = rows.length;
        if (row + col == n - 1) {
            antiDiagonal += toAdd;
        }

        if (Math.abs(rows[row]) == n || Math.abs(cols[col]) == n || Math.abs(diagonal) == n || 
            Math.abs(antiDiagonal) == n) {
            return player;
        }
        return 0;
    }

Follow Up:

1.（全职）略变形，设计数据结构，return的type是boolean，玩家一个是'W', 一个是‘Z’，要求考虑：第一，如果重复了之前的坐标怎么处理，第二，不能一个玩家连续走两次。他描述问题用了5分钟，20分钟写代码问follow-up。