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1275. Find Winner on a Tic Tac Toe Game

Difficulty Topics

Description

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

  • 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 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.

Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. 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 (i.e., 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, they always play first.

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.

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.

 

Constraints:

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

Solution

find-winner-on-a-tic-tac-toe-game.py
class Solution:
    def tictactoe(self, moves: List[List[int]]) -> str:
        n = 3
        rows = [0] * n
        cols = [0] * n
        diag1 = diag2 = 0
        currPlayer = 1

        for x, y in moves:
            rows[x] += currPlayer
            cols[y] += currPlayer
            diag1 = diag1 + currPlayer if x == y else diag1
            diag2 = diag2 + currPlayer if x + y == n - 1 else diag2

            if abs(rows[x]) == n or abs(cols[y]) == n or abs(diag1) == n or abs(diag2) == n:
                return "A" if currPlayer == 1 else "B"

            currPlayer *= -1

        return "Draw" if len(moves) == n * n else "Pending"