794. Valid Tic-Tac-Toe State (Medium)

https://leetcode.com/problems/valid-tic-tac-toe-state/

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O".  The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

  • Players take turns placing characters into empty squares (" ").
  • The first player always places "X" characters, while the second player always places "O" characters.
  • "X" and "O" characters are always placed into empty squares, never filled ones.
  • The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
  • The game also ends if all squares are non-empty.
  • No more moves can be played if the game is over.
Example 1:
Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:
Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:
Input: board = ["XOX", "O O", "XOX"]
Output: true

Note:

  • board is a length-3 array of strings, where each string board[i] has length 3.
  • Each board[i][j] is a character in the set {" ", "X", "O"}.

Solutions

class Solution {

    // This is a nasty problem, too many special cases, skip it.

    public boolean validTicTacToe(String[] board) {
        int diff = count(board, 'X') - count(board, 'O');

        // Initially, count of X must be equal to or 1 more than O, otherwise not a
        // valid Tic-Tac-Toe board
        if (diff > 1 || diff < 0) {
            return false;
        }

        if (hasWon(board, 'X')) {
            return diff == 1;
        }

        if (hasWon(board, 'O')) {
            return diff == 0;
        }

        return true;
    }

    private boolean hasWon(String[] board, char player) {
        int size = board.length;

        // search the winner horizontally
        for (int i = 0; i < size; i++) {
            int count = 0;
            for (int j = 0; j < size && board[i].charAt(j) == player; j++) {
                count++;
            }

            if (count == size) {
                return true;
            }
        }

        // search the winner vertically
        for (int j = 0; j < size; j++) {
            int count = 0;
            for (int i = 0; i < size && board[i].charAt(j) == player; i++) {
                count++;
            }

            if (count == size) {
                return true;
            }
        }

        // search the winner diagonally, from left top to right bottom corner
        int count = 0;
        for (int i = 0; i < size && board[i].charAt(i) == player; i++) {
            count++;
        }

        if (count == size) {
            return true;
        }

        // search the winner diagonally, from left bottom to right top corner
        for (int i = 0; i < size && board[size-i-1].charAt(i) == player; i++) {
            count++;
        }

        if (count == size) {
            return true;
        }

        return false;
    }

    private int count(String[] board, char player) {
        int count = 0;

        for (String row : board) {
            for (char c : row.toCharArray()) {
                if (c != player) {
                    continue;
                }

                count++;
            }
        }

        return count;
    }
}

Incorrect Solutions

References

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