# Java Priority Queue | HackerRank Solution

Hello coders, today we are going to solve Java Priority Queue HackerRank 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

Contents

## Problem

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 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 <= 105
• 2 <= |name| <= 30

## Output

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

`````` 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

`````` Dan
Ashley
Shafaet
Maria``````

#### Explanation

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.

## Solution –Java Priority Queue

```import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
// import Priority Queue Collection Framework
import java.util.*;
/*
* Create the Student and Priorities classes here.
*/
class Student implements Comparable<Student>{
String name = new String();
double cgpa;
int id;
public Student(String name,double cgpa,int id)
{
this.name = name;
this.cgpa = cgpa;
this.id = id;
}
public String getName(){
return this.name;
}
public int compareTo(Student s)
{
if(cgpa == s.cgpa)
{
if(name.compareTo(s.name) == 0)
{
if(id == s.id)
return 0;
else if (id > s.id)
return 1;
else
return -1;
}
else
return name.compareTo(s.name);
}
else if(cgpa > s.cgpa)
return -1;
else
return 1;
}
}

class Priorities{
public ArrayList<Student> getStudents(List<String> events)
{
int n = events.size();
PriorityQueue<Student> pq = new PriorityQueue<Student>();
for(String i:events)
{
String[] s = new String;
s = i.split("\\s");
if(s.length>1)
{
}
else
{
pq.poll();
}
}
while(pq.size()>1)
{
System.out.println(pq.poll().name);
}
return new ArrayList<Student>(pq);
}
}

public class Solution {
private final static Scanner scan = new Scanner(System.in);
private final static Priorities priorities = new Priorities();

public static void main(String[] args) {
int totalEvents = Integer.parseInt(scan.nextLine());
List<String> events = new ArrayList<>();

while (totalEvents-- != 0) {
String event = scan.nextLine();
}

List<Student> students = priorities.getStudents(events);

if (students.isEmpty()) {
System.out.println("EMPTY");
} else {
for (Student st: students) {
System.out.println(st.getName());
}
}
}
}```

Disclaimer: The above Problem ( Java Priority Queue ) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.