348. Design Tic-Tac-Toe

Description

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

  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 are allowed.
  3. 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?

Solutions

Solution 1

Python Code
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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)

Java Code
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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);
 */