## Algorithm

Problem Name: Java Priority Queue

In this HackerRank Functions in Java programming problem solution,

In computer science, a priority queue is an abstract data type which is like a regular queue, but where additionally each element has a "priority" associated with it. In a priority queue, an element with high priority is served before an element with low priority. - Wikipedia

In this problem we will test your knowledge on Java Priority Queue.

There are a number of students in a school who wait to be served. Two types of events, ENTER and SERVED, can take place which are described below.

• ENTER: A student with some priority enters the queue to be served.
• SERVED: The student with the highest priority is served (removed) from the queue.

A unique id is assigned to each student entering the queue. The queue serves the students based on the following criteria (priority criteria):

1. The student having the highest Cumulative Grade Point Average (CGPA) is served first.
2. Any students having the same CGPA will be served by name in ascending case-sensitive alphabetical order.
3. Any students having the same CGPA and name will be served in ascending order of the id.

Create the following two classes:

• The Student class should implement:
• The constructor Student(int id, String name, double cgpa).
• The method int getID() to return the id of the student.
• The method String getName() to return the name of the student.
• The method double getCGPA() to return the CGPA of the student.
• The Priorities class should implement the method List<Student> getStudents(List<String> events) to process all the given events and return all the students yet to be served in the priority order.

Input Format

The first line contains an integer, n, describing the total number of events. Each of the n subsequent lines will be of the following two forms:

• ENTER name CGPA id: The student to be inserted into the priority queue.
• SERVED: The highest priority student in the queue was served.

The locked stub code in the editor reads the input and tests the correctness of the Student and Priorities classes implementation.

Constraints

• 2 <= n <= 1000
• 0 <= CGPA <= 4.00
• 1 <= id <= 10**5
• 2 <= name <= 30

Output Format

The locked stub code prints the names of the students yet to be served in the priority order. If there are no such student, then the code prints EMPTY.

Sample Input 0

12
ENTER John 3.75 50
ENTER Mark 3.8 24
ENTER Shafaet 3.7 35
SERVED
SERVED
ENTER Samiha 3.85 36
SERVED
ENTER Ashley 3.9 42
ENTER Maria 3.6 46
ENTER Anik 3.95 49
ENTER Dan 3.95 50
SERVED

Sample Output 0

Dan
Ashley
Shafaet
Maria

Explanation 0

In this case, the number of events is 12. Let the name of the queue be Q.

• John is added to Q. So, it contains (John, 3.75, 50).
• Mark is added to Q. So, it contains (John, 3.75, 50) and (Mark, 3.8, 24).
• Shafaet is added to Q. So, it contains (John, 3.75, 50), (Mark, 3.8, 24), and (Shafaet, 3.7, 35).
• Mark is served as he has the highest CGPA. So, Q contains (John, 3.75, 50) and (Shafaet, 3.7, 35).
• John is served next as he has the highest CGPA. So, Q contains (Shafaet, 3.7, 35).
• Samiha is added to Q. So, it contains (Shafaet, 3.7, 35) and (Samiha, 3.85, 36).
• Samiha is served as she has the highest CGPA. So, Q contains (Shafaet, 3.7, 35).
• Now, four more students are added to Q. So, it contains (Shafaet, 3.7, 35), (Ashley, 3.9, 42), (Maria, 3.6, 46), (Anik, 3.95, 49), and (Dan, 3.95, 50).
• Anik is served because though both Anil and Dan have the highest CGPA but Anik comes first when sorted in alphabetic order. So, Q contains (Dan, 3.95, 50), (Ashley, 3.9, 42), (Shafaet, 3.7, 35), and (Maria, 3.6, 46).

As all events are completed, the name of each of the remaining students is printed on a new line.

## Code Examples

### #1 Code Example with Java Programming

Code - Java Programming

import java.util.Scanner;
import java.util.ArrayDeque;
import java.util.HashMap;

// Why not use just a HashMap instead of both a HashMap and an ArrayDeque?
// Well, an ArrayDeque helps keep the ordering of elements. Although the
// elements are also in our HashMap, they is no ordering to the elements
// in the HashMap since it's just a set.

//  Time Complexity: O(n)
// Space Complexity: O(n)

public class test {
public static void main(String[] args) {
HashMap < Integer, Integer> map = new HashMap();
ArrayDeque deque     = new ArrayDeque();

Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int m = scan.nextInt();
int max = 0;

for (int i = 0; i  <  n; i++) {
/* Remove old value (if necessary) */
if (i >= m) {
int old = deque.removeFirst();
if (map.get(old) == 1) {
map.remove(old);
} else {
map.merge(old, -1, Integer::sum);
}
}

int num = scan.nextInt();
map.merge(num, 1, Integer::sum);

max = Math.max(max, map.size());

/* If all integers are unique, we have found our largest
possible answer, so we can break out of loop */
if (max == m) {
break;
}
}

scan.close();
System.out.println(max);
}
}
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