PLEASE READ THIS FIRST! _____________________Please write in java. Add comments but make the comments short. No need for really long comments. Please keep code very neat and simple , dont add anything unneccesary in the code if you weren't instructed to do so. If youve answered this before please dont copy and paste from a previous question! (Rewrite it in another way)..... Also be sure to read the instructions below carefully. The two files that are attached are screenshots of the textbook's MAX-HEAP Example. You will be creating a mini-heap from that example. But the Instructions below explains everything you'll be doing. NO THIS is not a homework assignment, it's practice work. Please type the code out so that I can copy and paste it into my IDE and see the output for my self but also still provide a screenshot of your output!INSTRUCTIONS FOR QUESTION BELOW!!_____________________________You may use the Java Libraries for solving this problem. We recommend using java.util.* The star “ * ” acts as a wildcard and allows you to use the entire java.util library. The heap presented in the screenshot is also known as a max-heap, in which each node is greater than or equal to any of it's children. A min-heap is a heap in which each node is less than or equal to any of its children. Min-heaps are often used to implement priority queues. Revise the Heap class in the screenshot to implement a min-heap. (The max heap example begins on the screenshot that says "Listings 23.9" at the top).Use the following logic:Write a main method that will accept 5 numbers, and put them in a min-heapRemove them from the min heap, printing one at a time as they are removed, to show that the list is sortedHint: The screenshotted textbook example is for a MAX heap; you must alter that example to reflect a MIN heap – think about it ! 454648495051525354555657585960616264656668697071727374757677 }int leftChildIndex 2 currentIndex + 1;int rightChildIndex = 2* currentIndex + 2;}// Find the maximum between two childrenif (leftChildIndex>-list.size())int maxIndex = leftChildIndex:if (rightChildIndexlist.size()) {if (list.get(max Index).compareTo(list.get(rightChildIndex)) < 0) {maxIndex - rightChildIndex;}// Swap if the current node is less than the maximumif (list.get(currentIndex).compareTo(break; // The tree is a heaplist.get(max Index)) < 0) {E temp = list.get(maxIndex);list.set(maxIndex, list.get(currentIndex));list.set(currentIndex, temp):currentIndex = maxIndex:}elsebreak; // The tree is a heapreturn removedobject;}/** Get the number of nodes in the tree "/public int getSize() {return list.size():23.6 Heap Sort 879compare two childrenswap with the larger child internal heap representationno-arg constructorconstructoradd a new objectappend the objectswap with parentheap nowremove the rootempty heaprootnew rootremove the lastadjust the treeLISTING 23.9 Heap.java1 public class Heap> {2 private java.util.ArrayList list = new java.util.ArrayListo():3456791011121314151617181920212223242526272829303132333435363738394041424344/** Create a default heap */public Heap() {/* Create a heap from an array of objects */public Heap (E[] objects) {for (int i = 0; i 0) {int parentIndex - (currentIndex - 1) / 2;// Swap if the current object is greater than its parentif (list.get(current Index) .compareTo(list.get(parent Index)) > 0) {E temp = list.get(currentIndex):list.set(current Index, list.get (parentIndex));list.set(parentIndex, temp):}elsebreak; // The tree is a heap nowcurrentIndex - parentIndex;/** Remove the root from the heap */public E remove() {if (list.size() == 0) return null;E renovedObject - list.get(0);list.set(0, list.get(list.size()-1));list.remove(list.size()-1):int current Index - 0:while (currentIndex < list.size()) {

import java.util.*;class Heap{// vector for storing heapprivate Vector<Integer> A; //…

Answered: You may use the Java Libraries for…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

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PLEASE READ THIS FIRST! _____________________

Please write in java. Add comments but make the comments short. No need for really long comments. Please keep code very neat and simple , dont add anything unneccesary in the code if you weren't instructed to do so. If youve answered this before please dont copy and paste from a previous question! (Rewrite it in another way)..... Also be sure to read the instructions below carefully. The two files that are attached are screenshots of the textbook's MAX-HEAP Example. You will be creating a mini-heap from that example. But the Instructions below explains everything you'll be doing. NO THIS is not a homework assignment, it's practice work. Please type the code out so that I can copy and paste it into my IDE and see the output for my self but also still provide a screenshot of your output!

INSTRUCTIONS FOR QUESTION BELOW!!_____________________________

You may use the Java Libraries for solving this problem. We recommend using java.util.* The star “ * ” acts as a wildcard and allows you to use the entire java.util library.

The heap presented in the screenshot is also known as a max-heap, in which each node is greater than or equal to any of it's children. A min-heap is a heap in which each node is less than or equal to any of its children. Min-heaps are often used to implement priority queues. Revise the Heap class in the screenshot to implement a min-heap. (The max heap example begins on the screenshot that says "Listings 23.9" at the top).

Use the following logic:

Write a main method that will accept 5 numbers, and put them in a min-heap
Remove them from the min heap, printing one at a time as they are removed, to show that the list is sorted

Hint: The screenshotted textbook example is for a MAX heap; you must alter that example to reflect a MIN heap – think about it !

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int leftChildIndex 2 currentIndex + 1;
int rightChildIndex = 2* currentIndex + 2;
}
// Find the maximum between two children
if (leftChildIndex>-list.size())
int maxIndex = leftChildIndex:
if (rightChildIndex
list.size()) {
if (list.get(max Index).compareTo(
list.get(rightChildIndex)) < 0) {
maxIndex - rightChildIndex;
}
// Swap if the current node is less than the maximum
if (list.get(currentIndex).compareTo(
break; // The tree is a heap
list.get(max Index)) < 0) {
E temp = list.get(maxIndex);
list.set(maxIndex, list.get(currentIndex));
list.set(currentIndex, temp):
currentIndex = maxIndex:
}
else
break; // The tree is a heap
return removedobject;
}
/** Get the number of nodes in the tree "/
public int getSize() {
return list.size():
23.6 Heap Sort 879
compare two children
swap with the larger child

internal heap representation
no-arg constructor
constructor
add a new object
append the object
swap with parent
heap now
remove the root
empty heap
root
new root
remove the last
adjust the tree
LISTING 23.9 Heap.java
1 public class Heap<E extends Comparable<E>> {
2 private java.util.ArrayList<E> list = new java.util.ArrayListo():
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/** Create a default heap */
public Heap() {
/* Create a heap from an array of objects */
public Heap (E[] objects) {
for (int i = 0; i <objects.length; i++)
add(objects[1]);
}
/** Add a new object into the heap /
public void add(E newObject) {
list.add(newObject); // Append to the heap
int currentIndex-list.size()-1; // The index of the last node
while (currentIndex > 0) {
int parentIndex - (currentIndex - 1) / 2;
// Swap if the current object is greater than its parent
if (list.get(current Index) .compareTo(
list.get(parent Index)) > 0) {
E temp = list.get(currentIndex):
list.set(current Index, list.get (parentIndex));
list.set(parentIndex, temp):
}
else
break; // The tree is a heap now
currentIndex - parentIndex;
/** Remove the root from the heap */
public E remove() {
if (list.size() == 0) return null;
E renovedObject - list.get(0);
list.set(0, list.get(list.size()-1));
list.remove(list.size()-1):
int current Index - 0:
while (currentIndex < list.size()) {

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