794. Valid Tic-Tac-Toe State

Problem

Given a Tic-Tac-Toe board 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 that 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 three 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 = ["XOX","O O","XOX"]
Output: true

  Constraints:

• board.length == 3

• board[i].length == 3

• board[i][j] is either 'X', 'O', or ' '.

Solution (Java)

class Solution {
    public boolean validTicTacToe(String[] board) {
        // X=1,O=-1,’ ’=0
        int sum = 0;
        int[] winsCol = new int[3];
        int[] winsDiag = new int[2];
        boolean xWin = false;
        boolean oWin = false;
        for (int i = 0; i < 3; i++) {
            String str = board[i];
            int rowSum = 0;
            for (int j = 0; j < 3; j++) {
                // char chr=str.toCharArray()[j];
                int intchr = 0;
                if (str.toCharArray()[j] == 'X') {
                    intchr = 1;
                }
                if (str.toCharArray()[j] == 'O') {
                    intchr = -1;
                }
                rowSum += intchr;
                winsCol[j] += intchr;
                if (i == 2 && winsCol[j] == 3) {
                    xWin = true;
                }
                if (i == 2 && winsCol[j] == -3) {
                    oWin = true;
                }
                if (Math.abs(i - j) != 1) {
                    if (i == j && i == 1) {
                        winsDiag[0] += intchr;
                        winsDiag[1] += intchr;
                    } else if (i == j) {
                        winsDiag[0] += intchr;
                    } else {
                        winsDiag[1] += intchr;
                    }
                }
                if (i == 2 && Math.max(winsDiag[0], winsDiag[1]) == 3) {
                    xWin = true;
                }
                if (i == 2 && Math.min(winsDiag[0], winsDiag[1]) == -3) {
                    oWin = true;
                }
            }
            if (rowSum == 3) {
                xWin = true;
            }
            if (rowSum == -3) {
                oWin = true;
            }
            sum += rowSum;
        }
        if (sum == 0 && !xWin) {
            return true;
        }
        return sum == 1 && !oWin;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).