# 348. Design Tic-Tac-Toe

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?

```cpp
class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n) {
        board.resize(n, vector<int>(n, 0));
    }
    
    /** 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. */
    int move(int row, int col, int player) { // time: O(n)
        board[row][col] = player;
        int i = 0, j = 0, n = board.size();
        // Check row
        for (j = 1; j < n; ++j) {
            if (board[row][j] != board[row][j - 1]) break;
        }
        if (j == n) return player;
        
        // Check column
        for (i = 1; i < n; ++i) {
            if (board[i][col] != board[i - 1][col]) break;
        }
        if (i == n) return player;
        
        // Check diagonal
        if (row == col) {
            for (i = 1; i < n; ++i) {
                if (board[i][i] != board[i - 1][i - 1]) break;
            }
            if (i == n) return player;
        }
        
        // Check anti-diagonal
        if (row + col == n - 1) {
            for (i = 1; i < n; ++i) {
                if (board[n - 1 - i][i] != board[n - i][i - 1]) break;
            }
            if (i == n) return player;
        }
        return 0; // no one wins
    }
private:
    vector<vector<int> > board;
};
```

```cpp
class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n) : rows(n), cols(n), diag(0), anti_diag(0), 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. */
    int move(int row, int col, int player) {
        int add = (player == 1 ? 1 : -1);
        rows[row] += add;
        cols[col] += add;
        diag += (row == col) ? add : 0;
        anti_diag += (row + col == N - 1) ? add : 0;
        if (abs(rows[row]) == N || abs(cols[col]) == N || abs(diag) == N || abs(anti_diag) == N) return player;
        else return 0;
    }
private:
    vector<int> rows, cols;
    int diag, anti_diag, N;
};
```
