Find Winner on a Tic Tac Toe Game in Python

# Find Winner on a Tic Tac Toe Game in Python

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.``````

## Solution:

``````class Solution(object):
def tictactoe(moves):
def check(grid):
for x in range(3):
row = set([grid[x], grid[x], grid[x]])
if len(row) == 1 and grid[x] != 0:
return grid[x]

for x in range(3):
column = set([grid[x], grid[x], grid[x]])
if len(column) == 1 and grid[x] != 0:
return grid[x]

dg1 = set([grid, grid, grid])
dg2 = set([grid, grid, grid])
if len(dg1) == 1 or len(dg2) == 1 and grid != 0:
return grid

return 0

if not moves:
return ""
grid = [[0, 0, 0], [0, 0, 0], [0,0,0]]
user = 1
for x in moves:
grid[x][x] = user
if user == 1:
user = 2
else:
user = 1

result = check(grid)
if result == 1:
return "A"
elif result == 2:
return "B"
else:
if len(moves) == 9:
return "Draw"
else:
return "Pending"

input = [[0,0],[2,0],[1,1],[2,1],[2,2]]
print(Solution.tictactoe(input)) # A

input = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
print(Solution.tictactoe(input)) # B

input = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
print(Solution.tictactoe(input)) # Draw

input = [[0,0],[1,1]]
print(Solution.tictactoe(input)) # Pending``````