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Solution 7: Adversarial Grid Logic (Tic-Tac-Toe Design)

Approach Explanation

To check the win condition in O(1) time per move, we don't need a full grid. Instead, we track the sum of values for each row, column, and the two diagonals. Let Player 1 add 1 and Player 2 subtract 1. If any sum reaches n or -n, that player wins.

Step-by-Step Logic

  1. __init__: Create arrays rows and cols of size n, and two integers diag and anti_diag.
  2. move(row, col, player):
    • val = 1 if player == 1 else -1.
    • rows[row] += val.
    • cols[col] += val.
    • If row == col, diag += val.
    • If row + col == n - 1, anti_diag += val.
    • If abs(rows[row]) == n or abs(cols[col]) == n or abs(diag) == n or abs(anti_diag) == n, return player.
    • Else, return 0.

Complexity

  • Time Complexity: O(1) per move.
  • Space Complexity: O(N).

Code

class TicTacToe:
    def __init__(self, n: int):
        self.n = n
        self.rows = [0] * n
        self.cols = [0] * n
        self.diag = 0
        self.anti_diag = 0

    def move(self, row: int, col: int, player: int) -> int:
        val = 1 if player == 1 else -1
        
        self.rows[row] += val
        self.cols[col] += val
        if row == col:
            self.diag += val
        if row + col == self.n - 1:
            self.anti_diag += val
            
        if (abs(self.rows[row]) == self.n or 
            abs(self.cols[col]) == self.n or 
            abs(self.diag) == self.n or 
            abs(self.anti_diag) == self.n):
            return player
            
        return 0