Find winner on a tic-tac-toe game¶
Time: O(1); Space: O(1); easy
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 ” ” “
Notes:
1 <= len(moves)<= 9
len(moves[i]) == 2
0 <= moves[i][j] <= 2
There are no repeated elements on moves.
moves follow the rules of tic tac toe.
Hints:
It’s straightforward to check if A or B won or not, check for each row/column/diag if all the three are the same.
Then if no one wins, the game is a draw iff the board is full, i.e. moves.length = 9 otherwise is pending.
[1]:
class Solution1(object):
def tictactoe(self, moves):
"""
:type moves: List[List[int]]
:rtype: str
"""
row, col = [[0]*3 for _ in range(2)], [[0]*3 for _ in range(2)]
diag, anti_diag = [0]*2, [0]*2
p = 0
for r, c in moves:
row[p][r] += 1
col[p][c] += 1
diag[p] += r == c
anti_diag[p] += r+c == 2
if 3 in (row[p][r], col[p][c], diag[p], anti_diag[p]):
return "AB"[p]
p ^= 1
return "Draw" if len(moves) == 9 else "Pending"
[2]:
s = Solution1()
moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
assert s.tictactoe(moves) == "A"
moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
assert s.tictactoe(moves) == "B"
moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
assert s.tictactoe(moves) == "Draw"
moves = [[0,0],[1,1]]
assert s.tictactoe(moves) == "Pending"