"""
Solution for 1275. Find Winner on a Tic Tac Toe Game
https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/
"""
from typing import List

class Solution:
  """
  Runtime: 24 ms, faster than 95.22% of Python3 online submissions for Find Winner on a Tic Tac Toe Game.
  Memory Usage: 12.6 MB, less than 100.00% of Python3 online submissions for Find Winner on a Tic Tac Toe Game.
  """

  def first_solution(self, moves: List[List[int]]) -> str:
    board = [[''] * 3 for _ in range(3)]
    for curr, (i, j) in enumerate(moves):
      board[i][j] = 'A' if curr % 2 == 0 else 'B'
    for i in range(3):
      if all(board[i][j] == board[i][0] and board[i][j] for j in range(3)):
        return board[i][j]
      if all(board[j][i] == board[0][i] and board[j][i] for j in range(3)):
        return board[j][i]
    if all(board[i][i] == board[0][0] and board[i][i] for i in range(3)):
      return board[0][0]
    if all(board[i][3 - i - 1] == board[0][-1] and board[i][3 - i - 1] for i in
           range(3)):
      return board[0][-1]
    return 'Draw' if len(moves) == 9 else 'Pending'

  def second_solution(self, moves: List[List[int]]) -> str:
    rows, cols = [[0] * 3 for _ in range(2)], [[0] * 3 for _ in range(2)]
    d1, d2, player = [0] * 2, [0] * 2, 0
    for r, c in moves:
      rows[player][r] += 1
      cols[player][c] += 1
      d1[player] += r == c
      d2[player] += r + c == 2
      if 3 in (rows[player][r], cols[player][c], d1[player], d2[player]):
        return 'A' if player == 0 else 'B'
      player ^= 1
    if len(moves) == 9:
      return 'Draw'
    return 'Pending'

  def tictactoe(self, moves: List[List[int]]) -> str:
    """
    Tic-tac-toe is played by two players A and B on a 3 x 3 grid.

    Here are the rules of Tic-Tac-Toe:

    Players take turns placing characters into empty squares (" ").
    The first player A always places "X" characters, while the second player B always places "O" characters.
    "X" and "O" characters are always placed into empty squares, never on 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.
    Given an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.

    Return the winner of the game if it exists (A or B), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".

    You can assume that moves is valid (It follows the rules of Tic-Tac-Toe), the grid is initially empty and A will play first.



    Example 1:

    Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
    Output: "A"
    Explanation: "A" wins, he always plays first.
    "X  "    "X  "    "X  "    "X  "    "X  "
    "   " -> "   " -> " X " -> " X " -> " X "
    "   "    "O  "    "O  "    "OO "    "OOX"
    Example 2:

    Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
    Output: "B"
    Explanation: "B" wins.
    "X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"
    "   " -> " O " -> " O " -> " O " -> "XO " -> "XO "
    "   "    "   "    "   "    "   "    "   "    "O  "
    Example 3:

    Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
    Output: "Draw"
    Explanation: The game ends in a draw since there are no moves to make.
    "XXO"
    "OOX"
    "XOX"
    Example 4:

    Input: moves = [[0,0],[1,1]]
    Output: "Pending"
    Explanation: The game has not finished yet.
    "X  "
    " O "
    "   "


    Constraints:

    1 <= moves.length <= 9
    moves[i].length == 2
    0 <= moves[i][j] <= 2
    There are no repeated elements on moves.
    moves follow the rules of tic tac toe.

    Args:
      moves:

    Returns:

    """
    return self.second_solution(moves)

  """
  Simulation
  T: O(N), S: O(N)
  1. Initialize a matrix
  2. Loop through the array
    2.1 If the i is even, check the matrix[i][j] as 1
    2.2 If the i is odd, check the matrix[i][j] as 1
  3. Check if there is any row or column with same numbers
    3.1 If there is a row or column with the same numbers, then return the number
  4. Return "Draw", if the len(moves) == 9, otherwise "Pending"
  """