## 348. Design Tic-Tac-Toe

Design a Tic-tac-toe game that is played between two players on anxngrid.

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|
``````

Could you do better than O(n2) per`move()`operation?

Thoughts:

1. only need to record row, col, diagonal, antidiagonal: for each move, player1 add 1 and player 2 mimus 1

Code: T: O(n)

``````class TicTacToe {
public:
/** Initialize your data structure here. */
TicTacToe(int n) {
size = n;
rows.resize(n, 0);
cols.resize(n, 0);
diag = 0, anti_diag = 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) {
if(player == 1){
rows[row]++; cols[col]++;
if(row == col) diag++;
if(row + col == size - 1) anti_diag++;
if(rows[row] == size || cols[col] == size || diag == size || anti_diag == size) return 1;
}
else { //player 2
rows[row] --; cols[col]--;
if(row == col) diag--;
if(row + col == size - 1) anti_diag--;
if(rows[row] == -size || cols[col] == -size || diag == -size || anti_diag == -size) return 2;
}

return 0;
}

private:
vector<int> rows, cols;
int diag, anti_diag;
int size;
};

/**
* Your TicTacToe object will be instantiated and called as such:
* TicTacToe obj = new TicTacToe(n);
* int param_1 = obj.move(row,col,player);
*/
``````

Similar idea in Java

``````public class TicTacToe {
private int[] rows;
private int[] cols;
private int diagonal;
private int antiDiagonal;

/** Initialize your data structure here. */
public TicTacToe(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;

if (row == col)
{
}

if (col == (cols.length - row - 1))
{
}

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

return 0;
}
}
``````