794. Valid Tic-Tac-Toe State

Description

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 ' '.

Solutions

class Solution {
    private String[] board;

    public boolean validTicTacToe(String[] board) {
        this.board = board;
        int x = count('X'), o = count('O');
        if (x != o && x - 1 != o) {
            return false;
        }
        if (win('X') && x - 1 != o) {
            return false;
        }
        return !(win('O') && x != o);
    }

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

    private int count(char x) {
        int cnt = 0;
        for (var row : board) {
            for (var c : row.toCharArray()) {
                if (c == x) {
                    ++cnt;
                }
            }
        }
        return cnt;
    }
}

class Solution {
public:
    bool validTicTacToe(vector<string>& board) {
        auto count = [&](char x) {
            int ans = 0;
            for (auto& row : board)
                for (auto& c : row) ans += c == x;
            return ans;
        };
        auto win = [&](char x) {
            for (int i = 0; i < 3; ++i) {
                if (board[i][0] == x && board[i][1] == x && board[i][2] == x) return true;
                if (board[0][i] == x && board[1][i] == x && board[2][i] == x) return true;
            }
            if (board[0][0] == x && board[1][1] == x && board[2][2] == x) return true;
            return board[0][2] == x && board[1][1] == x && board[2][0] == x;
        };
        int x = count('X'), o = count('O');
        if (x != o && x - 1 != o) return false;
        if (win('X') && x - 1 != o) return false;
        return !(win('O') && x != o);
    }
};

class Solution:
    def validTicTacToe(self, board: List[str]) -> bool:
        def win(x):
            for i in range(3):
                if all(board[i][j] == x for j in range(3)):
                    return True
                if all(board[j][i] == x for j in range(3)):
                    return True
            if all(board[i][i] == x for i in range(3)):
                return True
            return all(board[i][2 - i] == x for i in range(3))

        x = sum(board[i][j] == 'X' for i in range(3) for j in range(3))
        o = sum(board[i][j] == 'O' for i in range(3) for j in range(3))
        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)

func validTicTacToe(board []string) bool {
	var x, o int
	for _, row := range board {
		for _, c := range row {
			if c == 'X' {
				x++
			} else if c == 'O' {
				o++
			}
		}
	}
	win := func(x byte) bool {
		for i := 0; i < 3; i++ {
			if board[i][0] == x && board[i][1] == x && board[i][2] == x {
				return true
			}
			if board[0][i] == x && board[1][i] == x && board[2][i] == x {
				return true
			}
		}
		if board[0][0] == x && board[1][1] == x && board[2][2] == x {
			return true
		}
		return board[0][2] == x && board[1][1] == x && board[2][0] == x
	}
	if x != o && x-1 != o {
		return false
	}
	if win('X') && x-1 != o {
		return false
	}
	return !(win('O') && x != o)
}

/**
 * @param {string[]} board
 * @return {boolean}
 */
var validTicTacToe = function (board) {
    function count(x) {
        let cnt = 0;
        for (const row of board) {
            for (const c of row) {
                cnt += c == x;
            }
        }
        return cnt;
    }
    function win(x) {
        for (let i = 0; i < 3; ++i) {
            if (board[i][0] == x && board[i][1] == x && board[i][2] == x) {
                return true;
            }
            if (board[0][i] == x && board[1][i] == x && board[2][i] == x) {
                return true;
            }
        }
        if (board[0][0] == x && board[1][1] == x && board[2][2] == x) {
            return true;
        }
        return board[0][2] == x && board[1][1] == x && board[2][0] == x;
    }
    const [x, o] = [count('X'), count('O')];
    if (x != o && x - 1 != o) {
        return false;
    }
    if (win('X') && x - 1 != o) {
        return false;
    }
    return !(win('O') && x != o);
};

794 - Valid Tic-Tac-Toe State

794. Valid Tic-Tac-Toe State

Description

Solutions

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794. Valid Tic-Tac-Toe State

Description

Solutions

All Problems

All Solutions