348. Design Tic-Tac-Toe

Problem description:

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
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?

Solution:

If a player wins a Tic-Tac-Toe, it will achieve one of the following conditions:

1. Totally n move on the same row.
2. Totally n move on the same column.
3. Totally n move on diagonal line.
4. Totally n move on anti-diagonal line.

Therefore, we can use two 1D array and two variable to solve this question.
For each move, add the value to rows, columns, diagonal and anti-diagonal. If at any time a row or column matches the size of the board then that player has won.

class TicTacToe:

    def __init__(self, n: int):
        # the idea is whenever a player have n moves in a row, col or diagonal wins
        self.row, self.col = [0]*n, [0]*n
        self.dia, self.antidia, self.n = 0, 0, n

    def move(self, row: int, col: int, player: int) -> int:
        toAdd = 1 if player == 1 else -1
        self.row[row] += toAdd
        self.col[col] += toAdd
        if row == col:
            self.dia += toAdd
        if row+col == self.n-1:
            self.antidia += toAdd
        
        if self.n in [abs(self.row[row]), abs(self.col[col]), abs(self.dia), abs(self.antidia)]:
            return player
        return 0

class TicTacToe {
public:
    vector<int> rows;
    vector<int> cols;
    int diagonal= 0;
    int antiDiagonal= 0;
    
    /** Initialize your data structure here. */
    TicTacToe(int n) {
        rows.resize(n, 0);
        cols.resize(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) {
        int toAdd= player == 1? 1: -1;
        rows[row]+= toAdd;
        cols[col]+= toAdd;
        if(row == col)
            diagonal+= toAdd;
        if(col == cols.size()- row-1)
            antiDiagonal+= toAdd;
        
        if(abs(rows[row]) == rows.size() ||
           abs(cols[col]) == rows.size() ||
           abs(diagonal) == rows.size() ||
           abs(antiDiagonal) == rows.size())
            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);
 */

time complexity: $O(n)$
space complexity: $O(n)$
reference:
https://goo.gl/YR2JYd