# LeetCode – Tic-Tac-Toe (Java)

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

Java Solution 1 - Naive

We can simply check the row, column and the diagonals and see if there is a winner.

```public class TicTacToe {   int[][] matrix;   /** Initialize your data structure here. */ public TicTacToe(int n) { matrix = new int[n][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) { matrix[row][col]=player;   //check row boolean win=true; for(int i=0; i<matrix.length; i++){ if(matrix[row][i]!=player){ win=false; break; } }   if(win) return player;   //check column win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][col]!=player){ win=false; break; } }   if(win) return player;   //check back diagonal win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][i]!=player){ win=false; break; } }   if(win) return player;   //check forward diagonal win=true; for(int i=0; i<matrix.length; i++){ if(matrix[i][matrix.length-i-1]!=player){ win=false; break; } }   if(win) return player;   return 0; } }```

Java Solution 2

```public class TicTacToe { int[] rows; int[] cols; int dc1; int dc2; int n; /** Initialize your data structure here. */ public TicTacToe(int n) { this.n=n; this.rows=new int[n]; this.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 val = (player==1?1:-1);   rows[row]+=val; cols[col]+=val;   if(row==col){ dc1+=val; } if(col==n-row-1){ dc2+=val; }   if(Math.abs(rows[row])==n || Math.abs(cols[col])==n || Math.abs(dc1)==n || Math.abs(dc2)==n){ return player; }   return 0; } }```
Category >> Algorithms
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• Yi-jhe Huang

The actual question here:

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|