Design Tic-Tac-Toe

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?

Hint

Could you trade extra space such that move() operation can be done in O(1)? You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

Train of Thought

Initially, I had not read the Hint in the question and came up with an O(n) solution. After reading the extremely helpful hint; a much easier approach became apparent. The key observation is that in order to win Tic-Tac-Toe you must have the entire row or column. Thus, we don't need to keep track of an entire n^2 board. We only need to keep a count for each row and column. If at any time a row or column matches the size of the board then that player has won.

To keep track of which player, I add one for Player1 and -1 for Player2. There are two additional variables to keep track of the count of the diagonals. Each time a player places a piece we just need to check the count of that row, column, diagonal and anti-diagonal.

Also see a very similar answer that I believe had beaten me to the punch. We came up with our solutions independently but they are very similar in principle.

Code

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;

rows[row] += toAdd;
cols[col] += toAdd;
if (row == col)
{
    diagonal += toAdd;
}

if (col == (cols.length - row - 1))
{
    antiDiagonal += toAdd;
}

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;

} }

Complexity

time : o(1) space : o(n)

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