human
I found an interesting problem on https://leetcode.com/problems/chalkboard-xor-game/:
We are given non-negative integers nums[i] which are written on a chalkboard.  Alice and Bob take turns erasing exactly one number from the chalkboard, with Alice starting first.  If erasing a number causes the bitwise XOR of all the elements of the chalkboard to become 0, then that player loses.  (Also, we'll say the bitwise XOR of one element is that element itself, and the bitwise XOR of no elements is 0.)

Also, if any player starts their turn with the bitwise XOR of all the elements of the chalkboard equal to 0, then that player wins.

Return True if and only if Alice wins the game, assuming both players play optimally.

Example:
Input: nums = [1, 1, 2]
Output: false
Explanation: 
Alice has two choices: erase 1 or erase 2. 
If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose. 
If Alice erases 2 first, now nums becomes [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.


Notes: 

       1 <= N <= 1000. 
       0 <= nums[i] <= 2^16.

I tried it in Python, but could not do it. Can you solve it?

 ... What is a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
     def validTicTacToe(self, board):
         """
         :type board: List[str]
         :rtype: bool
         """
         O, X = 0, 0
         for r in board:
             for c in r:
                 if c == 'O':
                     O += 1
                 elif c == 'X':
                     X += 1
         if O - X > 0:
             return False
         if X - O > 1:
             return False
         if X - O == 0:
             for r in board:
                 if r == 'XXX':
                     return False
             for c in range(2):
                 if board[0][c] == board[1][c] == board[2][c] == 'X':
                     return False
             if board[0][0] == board[1][1] == board[2][2] == 'X':
                 return False
             if board[2][0] == board[1][1] == board[0][2] == 'X':
                 return False
         else:
             for r in board:
                 if r == 'OOO':
                     return False
             for c in range(2):
                 if board[0][c] == board[1][c] == board[2][c] == 'O':
                     return False
             if board[0][0] == board[1][1] == board[2][2] == 'O':
                 return False
             if board[2][0] == board[1][1] == board[0][2] == 'O':
                 return False
         
         return True