Question

Formatted question description: https://leetcode.ca/all/348.html

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves are allowed.
A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

TicTacToe(int n) Initializes the object the size of the board n.
int move(int row, int col, int player) Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move, and the two players alternate in making moves. Return
- 0 if there is no winner after the move,
- 1 if player 1 is the winner after the move, or
- 2 if player 2 is the winner after the move.

Example 1:

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|

Constraints:

2 <= n <= 100
player is 1 or 2.
0 <= row, col < n
(row, col) are unique for each different call to move.
At most n² calls will be made to move.

Follow-up: Could you do better than O(n²) per move() operation?

Algorithm

The straightforward idea is, to create a board of size nxn, where 0 means that there is no pawn at that position, 1 means that player 1 puts the stone, and 2 means player 2.

Then every time a piece is added to the board, we scan the current row, column, diagonal, and reverse diagonal to see if there are three pieces connected, if there are three pieces, return the corresponding player, if not, return 0.

But, we can do better.

Create a one-dimensional array rows and cols of size n, as well as variable diagonal diag and inverse diagonal rev_diag.

The idea of this method is,

If player 1 puts a child in a certain column of the first row, then rows[0] will increment by 1,
If player 2 puts a child in a certain column of the first row, rows[0] will decrement by 1,

Then only when rows[0] is equal to n or -n, it means that the children in the first row are all placed by a player, and then the game ends and returns to that player. The other rows and columns, the diagonal and the inverse diagonal are all in this way,

Code

public class Design_Tic_Tac_Toe {
    int[][] matrix;

    /**
     * Initialize your data structure here.
     */
    public Design_Tic_Tac_Toe(int n) {
        matrix = 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) {

        int[] rows = new int[3];
        int[] cols = new int[3];
        int[] diag = new int[2];

        int val = player == 1 ? 1: -1;

        rows[row] += val; // assumption: valid input, no override for same location
        cols[col] += val;

        if (row == col) {
            diag[0] += val;
        }
        if (row + col == 2) {
            diag[1] += val;
        }

        if (Math.abs(rows[row]) == 3
            || Math.abs(cols[col]) == 3
            || Math.abs(diag[0]) == 3
            || Math.abs(diag[1]) == 3) {
            return player;
        }

        return 0;
    }

    public int move_naive_solution(int row, int col, int player) {

        matrix[row][col] = player;

        //check row
        boolean win = true;
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[row][i] != player) {
                win = false;
                break;
            }
        }
        if (win) return player;
        //check column
        win = true;
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[i][col] != player) {
                win = false;
                break;
            }
        }
        if (win) return player;
        //check back diagonal
        win = true;
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[i][i] != player) {
                win = false;
                break;
            }
        }
        if (win) return player;
        //check forward diagonal
        win = true;
        for (int i = 0; i < matrix.length; i++) {
            if (matrix[i][matrix.length - i - 1] != player) {
                win = false;
                break;
            }
        }
        if (win) return player;
        return 0;
    }
}

############

class TicTacToe {
    private int n;
    private int[][] counter;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        counter = new int[2][(n << 1) + 2];
        this.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) {
        counter[player - 1][row] += 1;
        counter[player - 1][col + n] += 1;
        if (row == col) {
            counter[player - 1][n << 1] += 1;
        }
        if (row + col == n - 1) {
            counter[player - 1][(n << 1) + 1] += 1;
        }
        if (counter[player - 1][row] == n || counter[player - 1][col + n] == n
            || counter[player - 1][n << 1] == n || counter[player - 1][(n << 1) + 1] == n) {
            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);
 */

// OJ: https://leetcode.com/problems/design-tic-tac-toe/
// Time: O(N)
// Space: O(N^2)
class TicTacToe {
private:
    vector<vector<int>> board;
    int n;
    int status(int row, int col) {
        int p = board[row][col];
        bool win = true;
        for (int i = 0; i < n && win; ++i) {
            if (board[row][i] != p) win = false;
        }
        if (win) return p;
        win = true;
        for (int i = 0; i < n && win; ++i) {
            if (board[i][col] != p) win = false;
        }
        if (win) return p;
        if (row == col) {
            win = true;
            for (int i = 0; i < n && win; ++i) {
                if (board[i][i] != p) win = false;
            }
            if (win) return p;
        }
        if (row + col == n - 1) {
            win = true;
            for (int i = 0; i < n && win; ++i) {
                if (board[i][n - 1 - i] != p) win = false;
            }
            if (win) return p;
        }
        return 0;
    }
public:
    TicTacToe(int n) : n(n) {
        board = vector<vector<int>>(n, vector<int>(n));
    }
    int move(int row, int col, int player) {
        board[row][col] = player;
        return status(row, col);
    }
};

class TicTacToe:
    def __init__(self, n: int):
        """
        Initialize your data structure here.
        """
        self.n = n
        self.counter = [[0] * ((n << 1) + 2) for _ in range(2)]

    def move(self, row: int, col: int, player: int) -> int:
        """
        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.
        """
        n = self.n
        self.counter[player - 1][row] += 1
        self.counter[player - 1][col + n] += 1
        if row == col:
            self.counter[player - 1][n << 1] += 1
        if row + col == n - 1:
            self.counter[player - 1][(n << 1) + 1] += 1
        if (
            self.counter[player - 1][row] == n
            or self.counter[player - 1][col + n] == n
            or self.counter[player - 1][n << 1] == n
            or self.counter[player - 1][(n << 1) + 1] == n
        ):
            return player
        return 0


# Your TicTacToe object will be instantiated and called as such:
# obj = TicTacToe(n)
# param_1 = obj.move(row,col,player)

############

class TicTacToe(object):
  def __init__(self, n):
    self.rows = [0] * n
    self.cols = [0] * n
    self.diag = self.antiDiag = 0
    self.n = n

    def move(row, col, player):
      delta = 3 - player * 2
      self.rows[row] += delta
      self.cols[col] += delta
      self.diag += row == col and delta
      self.antiDiag += row + col == self.n - 1 and delta
      if delta * self.n in [self.rows[row], self.cols[col], self.diag, self.antiDiag]:
        return player
      return 0

    self.move = move

# Your TicTacToe object will be instantiated and called as such:
# obj = TicTacToe(n)
# param_1 = obj.move(row,col,player)

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