Answer: B) Vertical Partitioning
Explanation:
As, vertical partitioning is performed on Relation X, it is used for dividing the relation X vertically in columns and it involves creation of tables and columns. They also use some additional tables to store left out columns. We cannot partition the column without perform any modification of value of the column. It only relies on keeping the particular attributes of relation X.
Answer:
For question one, the first line It is Iteration, The second line is Comparator, the third line is none of these is correct.
Question two, the index based method for (a) is O(1) (b) O(1) (c) O(N) (d) O(N)
Explanation:
<em>Solution to the question</em>
Question 1:
The Iteration operation is required by the Iterable interface.
n application can indicate a specific way to order the elements of a SortedABList list by passing a(n) Comparator so that we can customize the sorting object to a constructor of the list class.
Suppose a list names contains 8 elements. A call to names.add(0, "Albert") results in: (e) None of these is correct
Question 2:
let us assume that the LBList is built on top of a Linked List and the ABList is built on top of an array:
(a) the add method index based is O(1) in the average case and the O(N) in the worst case
(b) The Index based set operation is O(1) since we can simply move to any index of an array of time constant.
(c) It is O(N) since index Of method needs to look for the whole array (based on worst case or average) to get the index
(d) It is O(N) since index Of method needs to find the whole linked list (on worst case or average ) to search the index.
Answer:
- food_category = {
- "Fruits": ["Apple", "Strawberries", "Orange"],
- "Vegetables": ["Carrots", "Peas", "Onions"],
- "Meats": ["Chicken", "Beef", "Mutton"]
- }
-
- for cat in food_category:
- output = cat + ": "
- for food in food_category[cat]:
- output += food + ", "
- print(output)
-
- answer = input("Do you want to add category or food (Y - Yes N - No): ")
-
- while(answer == "Y"):
- choice = input("New Category (N) or New Food (nf): ")
-
- if(choice == "N"):
- new_cat = input("Enter new category name: ")
- if(new_cat not in food_category):
- food_category[new_cat] = ""
- elif(choice == "nf"):
- cat = input("Enter category: ")
- new_food = input("Enter new food name: ")
- food_category[cat].append(new_food)
-
- answer = input("Do you want to add new category or food (Y - Yes N - No): ")
-
- print("All food categories: ")
- for cat in food_category:
- print(cat)
-
- selected_cat = input("Select a category to view the food list: ")
- output = ""
- for food in food_category[selected_cat]:
- output += food + ", "
- print(output)
Explanation:
The solution code is written in Python 3.
Firstly, create a food category dictionary data structure (Line 1 -5). Next, display all the food category and the food in each category (Line 7 - 11)
Next prompt user to feedback if they wish to add new category or new food (Line 13).
While the answer is yes prompt user to enter their choice, either category or food and use if else if statements to handle the choice made by the user. If the choice is new category, prompt user to enter new category name and check if the new category is already exist in the dictionary. If not, add that category to dictionary (Line 18 - 21).
If the choice is food, nf, prompt user to input food category and enter the new food name and add it to the food_category list (Line 22 - 25).
After that prompt user to enter if they wish to continue to add new data (Line 27).
After collecting all info from user, write a for loop to display all the categories (Line 29 - 31).
At last, prompt user to select a category and display all the food of the selected category (Line 33 - 37).
