Topic: Doubly Linked List Deque Implement the following problem in the main case 0 (see attached photo) Your algorithm for the hierarchy problem should follow this: Evaluate the final set of operations first given the set of rules. Then, do the remove operations. Finally, do the add operations. DO NOT MIND THE #include "dlldeque.h" it is already implemented only the case 0 needs to be solved #include <iostream>#include <cstring>#include "dlldeque.h"using namespace std;int main(int argc, char** argv) { DLLDeque* deque = new DLLDeque(); int test; cin >> test; switch (test) { case 0: // perform your Hierarchy implementation here // utilize the deque initialized, // initialize variables you need before switch // you can use the print() method to debug, but not the final_print() // do not modify from this point onwards deque->final_print(); break; case 1: deque->addFirst(50); deque->addLast(70); deque->addFirst(90); cout << deque->removeFirst() << endl; cout << deque->removeLast() << endl; cout << deque->removeLast() << endl; cout << deque->isEmpty() << endl; deque->final_print(); break; case 2: deque->addFirst(5); deque->addFirst(3); cout << deque->removeFirst() << endl; cout << deque->removeLast() << endl; deque->addLast(7); deque->addFirst(9); deque->addLast(4); cout << deque->removeLast() << endl; deque->addLast(6); deque->addFirst(8); cout << deque->removeFirst() << endl; cout << deque->size() << endl; deque->final_print(); break; case 3: deque->addFirst(19); deque->addFirst(47); deque->addFirst(67); cout << deque->removeLast() << endl; deque->addLast(17); cout << deque->removeFirst() << endl; cout << deque->removeFirst() << endl; deque->addLast(4); deque->addFirst(30); cout << deque->size() << endl; deque->final_print(); break; case 4: deque->addFirst(41); cout << deque->removeLast() << endl; deque->addLast(97); deque->addLast(6); deque->addFirst(11); cout << deque->removeLast() << endl; cout << deque->removeFirst() << endl; deque->addLast(37); deque->addFirst(38); deque->addFirst(40); cout << deque->removeLast() << endl; deque->addFirst(91); cout << deque->size() << endl; deque->final_print(); break; case 5: deque->addLast(2); cout << deque->removeFirst() << endl; deque->addFirst(48); deque->addFirst(22); deque->addLast(39); deque->addLast(81); deque->addFirst(77); cout << deque->removeFirst() << endl; cout << deque->removeFirst() << endl; deque->addFirst(34); deque->addFirst(25); cout << deque->removeLast() << endl; cout << deque->removeFirst() << endl; cout << deque->size() << endl; deque->final_print(); break; case 6: deque->addFirst(11); deque->addLast(19); deque->addLast(82); deque->addFirst(89); deque->addFirst(55); cout << deque->removeFirst() << endl; cout << deque->removeLast() << endl; deque->addFirst(50); cout << deque->removeLast() << endl; cout << deque->removeFirst() << endl; deque->addLast(19); deque->addFirst(15); cout << deque->size() << endl; deque->final_print(); break; } return 0;} In the Hierarchy Problem, you are given a set of numbers. Beforeadding a number, sum up the digits of the number and consider thefollowing:• Rule of 2: If the sum is divisible by 2, add the number to thefirst.• Rule of 3: If the sum is divisible by 3, remove the first and addthe number to the last.• Rule of 5: If the sum is divisible by 5, remove the last and addthe number to the first and the last.Rule of 7: If the sum is divisible by 7, remove both the first andthe last and add the number to the first.• Rule of X: If the sum does not fall on any one of the following,add the number to the last.Statute of Limitations: If a sum is true to more than one of thefollowing, only apply two of the following rules starting fromthe first mentioned rule. You can only remove and add at mostone number from each side of the deque so if two rulescombined tells you to remove the first twice as in Rule of 3 andRule of 7, only remove one; the similar is true for adding thenumber.• Gatekeeper: If you encounter a zero (0), stop the loop.Note: Be careful in removing from an empty deque. With thisproblem, you will have to consider improving your remove methodsfor this to only return O when you attempt to remove from an emptydeque. int main(int argc, char** argv) {DLLDeque* dequeint test;cin >> test;switch (test)= new DLLDeque ();case 0:// perform your Hierarchy implementation here// utilize the deque initialized,// initialize variables you need before switch// you can use the print() method to debug, but not the final_print()// do not modify from this point onwardsdeque->finalprint();break;

Deque (doubly ended queue) is a linear data structure(i.e. it stores data in a sequential manner),…

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Author:James Kurose, Keith Ross

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Chapter1: Computer Networks And The Internet

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Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Topic: Doubly Linked List Deque

Implement the following problem in the main case 0 (see attached photo)

Your algorithm for the hierarchy problem should follow this: Evaluate the final set of operations first given the set of rules. Then, do the remove operations. Finally, do the add operations.

DO NOT MIND THE #include "dlldeque.h" it is already implemented only the case 0 needs to be solved

#include <iostream>

#include <cstring>

#include "dlldeque.h"

using namespace std;

int main(int argc, char** argv) {

DLLDeque* deque = new DLLDeque();

int test;

cin >> test;

switch (test) {

case 0:

// perform your Hierarchy implementation here

// utilize the deque initialized,

// initialize variables you need before switch

// you can use the print() method to debug, but not the final_print()

// do not modify from this point onwards

deque->final_print();

break;

case 1:

deque->addFirst(50);

deque->addLast(70);

deque->addFirst(90);

cout << deque->removeFirst() << endl;

cout << deque->removeLast() << endl;

cout << deque->isEmpty() << endl;

deque->final_print();

break;

case 2:

deque->addFirst(5);

deque->addFirst(3);

cout << deque->removeFirst() << endl;

cout << deque->removeLast() << endl;

deque->addLast(7);

deque->addFirst(9);

deque->addLast(4);

cout << deque->removeLast() << endl;

deque->addLast(6);

deque->addFirst(8);

cout << deque->removeFirst() << endl;

cout << deque->size() << endl;

deque->final_print();

break;

case 3:

deque->addFirst(19);

deque->addFirst(47);

deque->addFirst(67);

cout << deque->removeLast() << endl;

deque->addLast(17);

cout << deque->removeFirst() << endl;

deque->addLast(4);

deque->addFirst(30);

cout << deque->size() << endl;

deque->final_print();

break;

case 4:

deque->addFirst(41);

cout << deque->removeLast() << endl;

deque->addLast(97);

deque->addLast(6);

deque->addFirst(11);

cout << deque->removeLast() << endl;

cout << deque->removeFirst() << endl;

deque->addLast(37);

deque->addFirst(38);

deque->addFirst(40);

cout << deque->removeLast() << endl;

deque->addFirst(91);

cout << deque->size() << endl;

deque->final_print();

break;

case 5:

deque->addLast(2);

cout << deque->removeFirst() << endl;

deque->addFirst(48);

deque->addFirst(22);

deque->addLast(39);

deque->addLast(81);

deque->addFirst(77);

cout << deque->removeFirst() << endl;

deque->addFirst(34);

deque->addFirst(25);

cout << deque->removeLast() << endl;

cout << deque->removeFirst() << endl;

cout << deque->size() << endl;

deque->final_print();

break;

case 6:

deque->addFirst(11);

deque->addLast(19);

deque->addLast(82);

deque->addFirst(89);

deque->addFirst(55);

cout << deque->removeFirst() << endl;

cout << deque->removeLast() << endl;

deque->addFirst(50);

cout << deque->removeLast() << endl;

cout << deque->removeFirst() << endl;

deque->addLast(19);

deque->addFirst(15);

cout << deque->size() << endl;

deque->final_print();

break;

}

return 0;

}

In the Hierarchy Problem, you are given a set of numbers. Before
adding a number, sum up the digits of the number and consider the
following:
• Rule of 2: If the sum is divisible by 2, add the number to the
first.
• Rule of 3: If the sum is divisible by 3, remove the first and add
the number to the last.
• Rule of 5: If the sum is divisible by 5, remove the last and add
the number to the first and the last.
Rule of 7: If the sum is divisible by 7, remove both the first and
the last and add the number to the first.
• Rule of X: If the sum does not fall on any one of the following,
add the number to the last.
Statute of Limitations: If a sum is true to more than one of the
following, only apply two of the following rules starting from
the first mentioned rule. You can only remove and add at most
one number from each side of the deque so if two rules
combined tells you to remove the first twice as in Rule of 3 and
Rule of 7, only remove one; the similar is true for adding the
number.
• Gatekeeper: If you encounter a zero (0), stop the loop.
Note: Be careful in removing from an empty deque. With this
problem, you will have to consider improving your remove methods
for this to only return O when you attempt to remove from an empty
deque.

int main(int argc, char** argv) {
DLLDeque* deque
int test;
cin >> test;
switch (test)
= new DLLDeque ();
case 0:
// perform your Hierarchy implementation here
// utilize the deque initialized,
// initialize variables you need before switch
// you can use the print() method to debug, but not the final_print()
// do not modify from this point onwards
deque->finalprint();
break;

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