348

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?

Companies:
Microsoft, Amazon, Facebook, Google

Related Topics:
Design

Similar Questions:

• Valid Tic-Tac-Toe State (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/design-tic-tac-toe/
// Author: github.com/lzl124631x
// 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);
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/design-tic-tac-toe/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(N)
class TicTacToe {
private:
    vector<int> rows, cols;
    int n, diag = 0, anti = 0;
public:
    TicTacToe(int n) : n(n), rows(n, 0), cols(n, 0) {}
    
    int move(int row, int col, int player) {
        int d = player == 1 ? 1 : -1;
        rows[row] += d;
        cols[col] += d;
        if (row == col) diag += d;
        if (row + col == n - 1) anti += d;
        if (abs(rows[row]) == n
           || abs(cols[col]) == n
           || abs(diag) == n
           || abs(anti) == n) return player;
        return 0;
    }
};