A university wants to schedule the classrooms for final exams. The attributes are given below:
1 answer:
Answer:
See explaination
Explanation:
Clearly the entities are as follows:-
Course
Section
Room
Test
The relationship among entities are as follows:-
Course has Section
Test is conducted for Course
Test is conducted in Section
Test is conducted in Room
Attributes of each entities are as follows:-
Course (CourseID, Name, Department)
Section (SectionID, Enrollment)
Room (RoomNumber, Capacity, Building)
Test (Time)
Section is a week entity as, there may be same sections for different courses, therefore section uses the primary key of course entity as foreign key.
Also entity Test is dependent upon the entities Room,Section and Course, therefore primary keys of these entities will be used as foreign key in the Test entity.
Check attachment for the ER diagram
You might be interested in
Answer:
Answer:
Here is the JAVA program:
import java.util.Scanner; //to accept input from user
public class Main { //class name
public static void main(String[] args) { //start of main method
Scanner input = new Scanner(System.in); //creates Scanner object
int integer = input.nextInt(); //declares and reads the integer indicating the number of words
String stringArray[] = new String[integer]; //creates array to hold strings
for (int i = 0; i < integer; i++) { //iterates through the array
stringArray[i] = input.next(); } //reads strings
int frequency[] = new int[integer]; //creates array for frequencies
int count; //declares variable to count frequency of each word
for (int i = 0; i < frequency.length; i++) { //iterates through the frequency array
count = 0; //initializes count to 0
for (int j = 0; j < frequency.length; j++) { //iterates through array
if (stringArray[i].equals(stringArray[j])) { //if element at ith index of stringArray is equal to element at jth index of stringArray
count++; } } //adds 1 to the count
frequency[i] = count; } //adds count to ith index of frequency array
for (int i = 0; i < stringArray.length; i++) { //iterates through the array
System.out.println(stringArray[i] + " " + frequency[i]); } } } //displays each word in stringArray and its corresponding frequency in frequency array
Explanation:
I will explain the program with an example:
let integer = 3
for (int i = 0; i < integer; i++) this loop is used to read input strings into stringArray.
At first iteration:
i = 0
0<3
stringArray[i] = input.next(); this reads a word and stores it to ith index of stringArray. Lets say user enters the string "hey" so hey is stored in stringArray[0]
i++ becomes i = 1
At second iteration:
i = 1
1<3
stringArray[i] = input.next(); this reads a word and stores it to ith index of stringArray. Lets say user enters the string "hi" so hi is stored in stringArray[1]
i++ becomes i = 2
At third iteration:
i = 2
2<3
stringArray[i] = input.next(); this reads a word and stores it to ith index of stringArray. Lets say user enters the string "hi" so hi is stored in stringArray[2]
i++ becomes i = 3
Now the loop breaks as i<integers evaluates to false.
Next the outer loop for (int i = 0; i < frequency.length; i++)
and inner loop for (int j = 0; j < frequency.length; j++) iterate through the array and if condition if (stringArray[i].equals(stringArray[j])) checks if any of the words at specified indices is equal. This is used in order to check if any word in the list comes more than once. So if any word occurs more than once in the array, then its count is increased each time and the count value is stored in frequency[] array to store the count/frequency of each word in the stringArray. Since hey occurs once so its count is set to 1 and hi occurs twice so its frequency is set to 2. So the output of the entire program is:
hey 1
hi 2
hi 2
The screenshot of the program along with its output with the given example in question is attached.
