Valid Tic-Tac-Toe State Python Solution

Tested Python solution for LeetCode 794 with 24 pytest cases. Generate a practice environment with lcpy.

LeetCode 794, Medium. Topics: Array, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 24 parametrized pytest cases, and a playground notebook:

lcpy gen -n 794   # by problem number
lcpy gen -s valid_tic_tac_toe_state   # by problem name

​

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.

​

Examples

Input: board = ["O  ","   ","   "]
Output: false
Explanation: The first player always plays "X".

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

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

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(1) - the board is always 3 x 3
    # Space: O(1)

    def valid_tic_tac_toe(self, board: list[str]) -> bool:
        x_count = sum(row.count("X") for row in board)
        o_count = sum(row.count("O") for row in board)

        if o_count not in (x_count - 1, x_count):
            return False

        x_wins = self._wins(board, "X")
        o_wins = self._wins(board, "O")

        if x_wins and o_wins:
            return False
        if x_wins:
            return x_count == o_count + 1
        if o_wins:
            return x_count == o_count
        return True

    def _wins(self, board: list[str], player: str) -> bool:
        lines = [board[i] for i in range(3)]
        lines += ["".join(row[j] for row in board) for j in range(3)]
        lines.append("".join(board[i][i] for i in range(3)))
        lines.append("".join(board[i][2 - i] for i in range(3)))
        return any(line == player * 3 for line in lines)

​

Complexity

Time	Space
O(1) - the board is always 3 x 3	O(1)

​

Tags