348. Design Tic-Tac-Toe
Problem:
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.
Solutions:
public class TicTacToe {
private int[][] board;
private int n;
private int rowSum[];
private int colSum[];
private int diag;
private int antiDiag;
public TicTacToe(int n) {
board = new int[n][n];
this.n = n;
rowSum = new int[n];
colSum = new int[n];
diag = 0;
antiDiag = 0;
}
public int move(int row, int col, int player) {
int add = 1;
if (player == 1) {
add = -1;
}
rowSum[row] += add;
colSum[col] += add;
if (row == col) {
diag += add;
}
if (row + col == n - 1) {
antiDiag += add;
}
if (rowSum[row] == n || colSum[col] == n || diag == n || antiDiag == n) {
return 2;
}
if (rowSum[row] == -n || colSum[col] == -n || diag == -n || antiDiag == -n) {
return 1;
}
return 0;
}
}