Massive Algorithms: LeetCode 348

LeetCode 348 - Design Tic-Tac-Toe

http://www.cnblogs.com/grandyang/p/5467118.html

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, 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.

Follow up中让我们用更高效的方法，那么根据提示中的，我们建立一个大小为n的一维数组rows和cols，还有变量对角线diag和逆对角线rev_diag，这种方法的思路是，如果玩家1在第一行某一列放了一个子，那么rows[0]自增1，如果玩家2在第一行某一列放了一个子，则rows[0]自减1，那么只有当rows[0]等于n或者-n的时候，表示第一行的子都是tne一个玩家放的，则游戏结束返回该玩家即可，其他各行各列，对角线和逆对角线都是这种思路，参见代码如下：
https://leetcode.com/discuss/101219/7-8-lines-o-1-java-python

public class TicTacToe {

    public TicTacToe(int n) {
        count = new int[6*n][3];
    }

    public int move(int row, int col, int player) {
        int n = count.length / 6;
        for (int x : new int[]{row, n+col, 2*n+row+col, 5*n+row-col})
            if (++count[x][player] == n)
                return player;
        return 0;
    }

    int[][] count;
}

https://leetcode.com/discuss/101717/java-o-1-short-and-clean

https://leetcode.com/discuss/101144/java-o-1-solution-easy-to-understand
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.

The key observation is that in order to win Tic-Tac-Toe you must have the entire row or column. Thus, we don't need to keep track of an entire n^2 board. We only need to keep a count for each row and column. If at any time a row or column matches the size of the board then that player has won.

To keep track of which player, I add one for Player1 and -1 for Player2. There are two additional variables to keep track of the count of the diagonals. Each time a player places a piece we just need to check the count of that row, column, diagonal and anti-diagonal.

private int[] rows; private int[] cols; private int diagonal; private int antiDiagonal; /** Initialize your data structure here. */ public TicTacToe(int n) { rows = new int[n]; 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 toAdd = player == 1 ? 1 : -1; rows[row] += toAdd; cols[col] += toAdd; if (row == col) { diagonal += toAdd; } if (col == (cols.length - row - 1)) { antiDiagonal += toAdd; } int size = rows.length; if (Math.abs(rows[row]) == size || Math.abs(cols[col]) == size || Math.abs(diagonal) == size || Math.abs(antiDiagonal) == size) { return player; } return 0; }
https://leetcode.com/discuss/101143/13-lines-simple-and-clean-o-1-java-solution
int[] rows, cols; int n, diagonal = 0, antiDiagonal = 0; public TicTacToe(int n) { this.n = n; rows = new int[n]; cols = new int[n]; } public int move(int row, int col, int player) { if(player == 1){ if(++rows[row] == n || ++cols[col] == n) return 1; if(row == col && ++diagonal == n) return 1; if(row + col == n - 1 && ++antiDiagonal == n) return 1; }else{ if(--rows[row] == -n || --cols[col] == -n) return 2; if(row == col && --diagonal == -n) return 2; if(row + col == n - 1 && --antiDiagonal == -n) return 2; } return 0; }

http://buttercola.blogspot.com/2016/06/leetcode-348-design-tic-tac-toe.html

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;

}

}

Not good - just different
http://dartmooryao.blogspot.com/2016/05/leetcode-348-design-tic-tac-toe.html
Map<String, Integer> map;
int n;

/** Initialize your data structure here. */
public TicTacToe(int n) {
this.n = n;
this.map = new HashMap<>();
}

/** 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) {
String rowKey = "R"+row+"_"+player;
int countR = map.getOrDefault(rowKey, 0)+1;
map.put(rowKey, countR);

String colKey = "C"+col+"_"+player;
int countC = map.getOrDefault(colKey, 0)+1;
map.put(colKey, countC);

int countD1 = 0;
if(row == col){
String d1Key = "D1_"+player;
countD1 = map.getOrDefault(d1Key, 0)+1;
map.put(d1Key, countD1);
}

int countD2 = 0;
if(row + col == n-1){
String d2Key = "D2_"+player;
countD2 = map.getOrDefault(d2Key, 0)+1;
map.put(d2Key, countD2);
}

if(countR == n || countC == n || countD1 == n || countD2 == n){
return player;
}else{
return 0;
}
}

O(n2)的解法，这种方法的思路很straightforward，就是建立一个nxn大小的board，其中0表示该位置没有棋子，1表示玩家1放的子，2表示玩家2。那么棋盘上每增加一个子，我们都每行每列，对角线和逆对角线来扫描一遍棋盘，看看是否有三子相连的情况，有的话则返回对应的玩家，没有则返回0

    TicTacToe(int n) {
        board.resize(n, vector<int>(n, 0));   
    }

    int move(int row, int col, int player) {
        board[row][col] = player;
        int i = 0, j = 0, N = board.size();
        for (i = 0; i < N; ++i) {
            if (board[i][0] != 0) {
                for (j = 1; j < N; ++j) {
                    if (board[i][j] != board[i][j - 1]) break;
                }
                if (j == N) return board[i][0];
            }
        }
        for (j = 0; j < N; ++j) {
            if (board[0][j] != 0) {
                for (i = 1; i < N; ++i) {
                    if (board[i][j] != board[i - 1][j]) break;
                }
                if (i == N) return board[0][j];
            }
        }
        if (board[0][0] != 0) {
            for (i = 1; i < N; ++i) {
                if (board[i][i] != board[i - 1][i - 1]) break;
            }
            if (i == N) return board[0][0];
        }
        if (board[N - 1][0] != 0) {
            for (i = 1; i < N; ++i) {
                if (board[N - i - 1][i] != board[N - i][i - 1]) break;
            }
            if (i == N) return board[N - 1][0];
        }
        return 0;
    }
    vector<vector<int>> board;

http://stackoverflow.com/questions/22087006/using-arrays-to-detect-a-win-in-a-gomoku-game
https://github.com/javierchavez/Gomoku/blob/master/Gomoku.java

LeetCode 348 - Design Tic-Tac-Toe

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