Question
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(n) per move() operation?
- Could you trade extra space such that move() operation can be done in O(1)?
Solution
TODO
Code
class TicTacToe {
int[] rows;
int[] cols;
int diagonal;
int antiDiagnal;
int n;
/** Initialize your data structure here. */
public TicTacToe(int n) {
this.n = 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 multiplier;
if(player == 1) multiplier = 1;
else multiplier = -1;
rows[row] += multiplier;
cols[col] += multiplier;
if(row == col) diagonal += multiplier;
if(row == n-col-1) antiDiagnal += multiplier;
int winScore = multiplier * n;
if(rows[row] == winScore ||
cols[col] == winScore ||
diagonal == winScore ||
antiDiagnal == winScore)
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);
*/
Performance
O(1) time complexity for move and O(n) space.