README_EN.md

794. Valid Tic-Tac-Toe State

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Description

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

Python3

class Solution:
    def validTicTacToe(self, board: List[str]) -> bool:

        def win(p):
            for i in range(3):
                if board[i][0] == board[i][1] == board[i][2] == p:
                    return True
                if board[0][i] == board[1][i] == board[2][i] == p:
                    return True
            if board[0][0] == board[1][1] == board[2][2] == p:
                return True
            return board[0][2] == board[1][1] == board[2][0] == p

        x, o = 0, 0
        for i in range(3):
            for j in range(3):
                if board[i][j] == 'X':
                    x += 1
                elif board[i][j] == 'O':
                    o += 1

        if x != o and x - 1 != o:
            return False

        if win('X') and x - 1 != o:
            return False

        return not (win('O') and x != o)

Java

class Solution {
    public boolean validTicTacToe(String[] board) {
        int x = 0, o = 0;
        for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (board[i].charAt(j) == 'X') {
                    x++;
                } else if (board[i].charAt(j) == 'O') {
                    o++;
                }
            }
        }
        if (x != o && x - 1 != o) {
            return false;
        }
        if (win(board, 'X') && x - 1 != o) {
            return false;
        }
        return !(win(board, 'O') && x != o);
    }

    private boolean win(String[] b, char p) {
        for (int i = 0; i < 3; i++) {
            if (b[i].charAt(0) == p && b[i].charAt(1) == p && b[i].charAt(2) == p) {
                return true;
            }
            if (b[0].charAt(i) == p && b[1].charAt(i) == p && b[2].charAt(i) == p) {
                return true;
            }
        }
        if (b[0].charAt(0) == p && b[1].charAt(1) == p && b[2].charAt(2) == p) {
            return true;
        }
        return b[0].charAt(2) == p && b[1].charAt(1) == p && b[2].charAt(0) == p;
    }
}

Go

func validTicTacToe(board []string) bool {
	x, o := 0, 0
	for i := 0; i < 3; i++ {
		for j := 0; j < 3; j++ {
			if board[i][j] == 'X' {
				x++
			} else if board[i][j] == 'O' {
				o++
			}
		}
	}
	if x != o && x-1 != o {
		return false
	}
	if win(board, 'X') && x-1 != o {
		return false
	}
	return !(win(board, 'O') && x != o)
}

func win(b []string, p byte) bool {
	for i := 0; i < 3; i++ {
		if b[i][0] == p && b[i][1] == p && b[i][2] == p {
			return true
		}
		if b[0][i] == p && b[1][i] == p && b[2][i] == p {
			return true
		}
	}
	if b[0][0] == p && b[1][1] == p && b[2][2] == p {
		return true
	}
	return b[0][2] == p && b[1][1] == p && b[2][0] == p
}

C++

class Solution {
public:
    bool validTicTacToe(vector<string>& board) {
        int x = 0, o = 0;
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (board[i][j] == 'X') {
                    ++x;
                } else if (board[i][j] == 'O') {
                    ++o;
                }
            }
        }
        if (x != o && x - 1 != o) {
            return false;
        }
        if (win(board, 'X') && x - 1 != o) {
            return false;
        }
        return !(win(board, 'O') && x != o);
    }

    bool win(vector<string>& b, char p) {
        for (int i = 0; i < 3; ++i) {
            if (b[i][0] == p && b[i][1] == p && b[i][2] == p) {
                return true;
            }
            if (b[0][i] == p && b[1][i] == p && b[2][i] == p) {
                return true;
            }
        }
        if (b[0][0] == p && b[1][1] == p && b[2][2] == p) {
            return true;
        }
        return b[0][2] == p && b[1][1] == p && b[2][0] == p;
    }
};

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