Thursday, June 2, 2016

Leetcode: 348. Design Tic-Tac-Toe

Design a Tic-tac-toe game that is played between two players on a n x n grid.
You may assume the following rules:
  1. A move is guaranteed to be valid and is placed on an empty block.
  2. Once a winning condition is reached, no more moves is allowed.
  3. 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:
  1. Could you trade extra space such that move() operation can be done in O(1)?
  2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.
Solution:
Use addtional arrays rows[n], cols[n] and two varialbes diagonal, anti_diagonal to mark the number of Xs and Os. 

Code (Java):
public class TicTacToe {
    private int[][] rows;
    private int[][] cols;
    private int[] diag;
    private int[] xdiag;
    private int n;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
        rows = new int[2][n];
        cols = new int[2][n];
        diag = new int[2];
        xdiag = new int[2];
    }
    
    /** 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 p = player == 1 ? 0 : 1;
        
        rows[p][row]++;
        cols[p][col]++;
        
        if (row == col) {
            diag[p]++;
        }
            
        // X-diagonal
        if (row + col == n - 1) {
            xdiag[p]++;
        }
            
        // If any of them equals to n, return 1
        if (rows[p][row] == n || cols[p][col] == n || 
            diag[p] == n || xdiag[p] == n) {
            return p + 1;
        }
        
        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);
 */


A Neat Code:
In the previous solution, we allocate two arrays for player 1 and 2, respectively. Actually we can use only one array for both of the players. Say, if it is player 1 put one chess, add that location by 1. If it is player 2, deduce it by one. Finally, if either player 1 or player 2 win, that location must be equal to n or -n. 

Code (Java):
public class TicTacToe {
    private int[] rows;
    private int[] cols;
    private int diag;
    private int xdiag;
    private int n;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
        rows = new int[n];
        cols = new int[n];
        diag = 0;
        xdiag = 0;
    }
    
    /** 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 count = player == 1 ? 1 : -1;
        
        rows[row] += count;
        cols[col] += count;
        
        if (row == col) {
            diag += count;
        }
            
        // X-diagonal
        if (row + col == n - 1) {
            xdiag += count;
        }
            
        // If any of them equals to n, return 1
        if (Math.abs(rows[row]) == n || 
            Math.abs(cols[col]) == n || 
            Math.abs(diag) == n || 
            Math.abs(xdiag) == n) {
            return count > 0 ? 1 : 2;
        }
        
        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);
 */

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