348. Design Tic-Tac-Toe

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348. Design Tic-Tac-Toe

Description

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

  • A move is guaranteed to be valid and is placed on an empty block.
  • Once a winning condition is reached, no more moves are allowed.
  • A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

  • TicTacToe(int n) Initializes the object the size of the board n.

  • int move(int row, int col, int player) Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move, and the two players alternate in making moves. Return

  • 0 if there is no winner after the move,

  • 1 if player 1 is the winner after the move, or

  • 2 if player 2 is the winner after the move.

Example 1:

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Input

["TicTacToe", "move", "move", "move", "move", "move", "move", "move"]
[[3], [0, 0, 1], [0, 2, 2], [2, 2, 1], [1, 1, 2], [2, 0, 1], [1, 0, 2], [2, 1, 1]]
Output

[null, 0, 0, 0, 0, 0, 0, 1]

Explanation

TicTacToe ticTacToe = new TicTacToe(3);
Assume that player 1 is "X" and player 2 is "O" in the board.
ticTacToe.move(0, 0, 1); // return 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

ticTacToe.move(0, 2, 2); // return 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

ticTacToe.move(2, 2, 1); // return 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

ticTacToe.move(1, 1, 2); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

ticTacToe.move(2, 0, 1); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

ticTacToe.move(1, 0, 2); // return 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

ticTacToe.move(2, 1, 1); // return 1(player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

Constraints:

  • 2 <= n <= 100
  • player is 1 or 2.
  • 0 <= row, col < n
  • (row, col) are unique for each different call to move.
  • At most n^2 calls will be made to move.

Follow-up: Could you do better than O(n^2) per move() operation?

Hints/Notes

  • 2025/01/20
  • set
  • premium

Solution

Language: C++

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class TicTacToe {
public:
    int diag1, diag2, n;
    vector<int> cols;
    vector<int> rows;
    TicTacToe(int n) {
        diag1 = diag2 = 0;
        this->n = n;
        cols.resize(n, 0);
        rows.resize(n, 0);
    }

    int move(int row, int col, int player) {
        int curPlayer = (player == 1) ? 1 : -1;
        if (row == col) {
            diag1 += curPlayer;
        }
        if (row + col == n - 1) {
            diag2 += curPlayer;
        }
        cols[col] += curPlayer;
        rows[row] += curPlayer;
        if (abs(diag1) == n || abs(diag2) == n ||
            abs(cols[col]) == n || abs(rows[row]) == n) {
            return player;
        }
        return 0;
    }
};

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe* obj = new TicTacToe(n);
 * int param_1 = obj->move(row,col,player);
 */