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Tic-Tac-Toe

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Problem Statement:

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

A move is guaranteed to be valid and is placed on an empty block. Once a winning condition is reached, no more moves are allowed. 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 player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

Example:

Assume that player 1 is "X" and player 2 is "O" in the board.
TicTacToe ticTacToe = new TicTacToe(3);

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|


Java and Python Solution:

This is a very simple yet interesting problem. While solutioning for this problem we will focus on how we go on optimizing our solution from O(n2) to all the way to O(1).

Brute Force O(n2) Algorithm:



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O(n) Algorithm:



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O(1) Algorithm:



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Instructor: