348. Design Tic-Tac-Toe

// LeetCode URL: https://leetcode.com/problems/design-tic-tac-toe/
/**
 * Time Complexity: Initialization -> O(1). Move -> O(1).
 *
 * Space Complexity: O(N). This is required to store scores at each row and each
 * column.
 */
class TicTacToe {

    int[] rows;
    int[] cols;
    int diag1;
    int diag2;
    int size;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        size = n;
        rows = new int[n];
        cols = new int[n];
        diag1 = 0;
        diag2 = 0;
    }

    /**
     * Player {player} makes a move at ({row}, {col}).
     *
     * @param row    The row of the board.
     * @param col    The column of the board.
     * @param player The player, can be either 1 or 2.
     * @return The current winning condition, can be either: 0: No one wins. 1:
     *         Player 1 wins. 2: Player 2 wins.
     */
    public int move(int row, int col, int player) {
        int inc = player == 1 ? 1 : -1;

        rows[row] += inc;
        cols[col] += inc;
        if (row == col) {
            diag1 += inc;
        }
        if (row + col == size - 1) {
            diag2 += inc;
        }

        if (Math.abs(rows[row]) == size || Math.abs(cols[col]) == size || Math.abs(diag1) == size
                || Math.abs(diag2) == size) {
            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);
 */