LeetCode – Tic-Tac-Toe (Java)

Category: Algorithms   May 20, 2014

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;
    }
}

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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|
    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.