348. Design Tic-Tac-Toe

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?

Solution:

class TicTacToe {
    int[][] board;
    int n;
    public TicTacToe(int n) {
        board = new int[n][n];
        this.n = n;
    }


    public int move(int row, int col, int player) {
        board[row][col] = player;
        // row
        for (int i = 0; i < n; i++){
            if (board[row][i] != player){
                break;
            }

            if (i == n -1){
                return player;
            }
        }
        // col
        for (int i = 0; i < n; i++){
            if (board[i][col] != player){
                break;
            }

            if (i == n -1){
                return player;
            }
        }
        // dia

        if (row == col){
            for (int i = 0; i < n; i++){
                if (board[i][i] != player){
                    break;
                }

                if (i == n - 1){
                    return player;
                }
            }
            // 0,2
            // row - col = 0 - 2 = -2
            // 1,1 = row 1 - 1 = 0
            // (2,0)
            // 0  0 0 
            // 0  0 0 
            // 0  0 0
            // row - col = 
            // 0 - (n - 1) = - n - 1
            // 1 
            // 2 . 
        }

        for (int i = n - 1; i >= 0; i--){
            if (board[n - 1 - i][i] != player){
                    // 2 - 2 = 0 , 2 
             //  2 - 1 = 1 , 1
                // 2, 0
                break;
            }

            if (i == 0){
                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);
 */

 /*
 0 1
 0 2
 */

 // TC: O(n)
 // SC: O(n^2)