Explain the flow of the code. See attached photo for the problem. (This is not graded)int removeAt(int pos) { if (pos <= index){ int elem = 0; if (pos == 1){ return removeHead(); } node* currnode = head; node* prevnode = NULL; int count = 0; while (currnode != NULL) { count++; if (count == pos) { elem = currnode-> element; prevnode->next = currnode->next; if (currnode == tail) { tail = prevnode; } free(currnode); index--; return elem; }else { prevnode = currnode; currnode = currnode->next; } } } return -1; } int removeAll(int num) { int count=0; node* currnode = head; node* prevnode = NULL; while (currnode != NULL) { if (currnode->element == num) { count++; if (prevnode != NULL) { if (currnode == tail) { tail = prevnode; } node* rem = currnode; prevnode->next = currnode->next; currnode = currnode->next; free(rem); index--; } else { node* rem = currnode; currnode = rem->next; removeHead(); } } else { prevnode = currnode; currnode = currnode->next; } } return count; } int contains(int num) { for (int i=1; i <= index; i++){ if (get(i) == num){ return i; } } return 0; } int count(int num) { int count =0; for (int i =1; i <= index; i++){ if (get(i) == num){ count++; } } return count; } bool move(int pos1, int pos2) { if ( pos1 <= index && pos2 <= index){ int elem = removeAt(pos1); if(pos2 == 1){ addHead(elem); } else{ addAt(elem,pos2); } return true; } return false; } int removeTail() { int elem=0; node* currnode = head; node* prevnode = NULL; while (currnode != NULL) { if (currnode == tail) { elem = currnode-> element; if (prevnode != NULL){ prevnode->next = currnode->next; } tail = prevnode; if(tail == NULL){ head = NULL; } free(currnode); index--; return elem; }else { prevnode = currnode; currnode = currnode->next; } } return 0; } bool isEmpty() { if (index == 0){ return true; } return false; } void clear() { while(head != NULL){ node* temp = head->next; removeHead(); head = temp; } } int count(int num)This will return the count of the instances of the element numin the list. In the linked list in removeAll method, having themethod count(10) will return 3 as there are three 10's in thelinked list.Additionally, you are to implement these methods:bool move(int pos1, int pos2)This method will move the element at position pos1 to theposition specified as pos2 and will return true if the move issuccessful. This will return false if either the specifiedpositions is invalid. In the example stated, operating themethod move(1, 3) will move the 1st element and shall nowbecome the 3rd element. The linked list shall now look like: 30→ 40 + 10 + 50• int removeTail()This method will remove the last element in your linked list andreturn the said element. If the list is empty, return 0. You canuse the existing methods.• void isEmpty)This method will return true if the linked list is empty,otherwise return false.• void clear()This method will empty your linked list. Effectively, this shouldand already has been called in your destructor (ie., the-LinkedList() method) so that it will deallocate the nodescreated first before deallocating the linked list itself. • int removeAt(int pos)Removes the number in the posth position of the list andreturns the element removed.Performing removeAt(3) in the example list will remove the 3rdelement of the linked list and the updated list will be: 10 → 30 →50When the value of pos is greater than the size or less than one,return -1.• int removeAll(int num)Removes all instances of num in the linked list and returns thenumber of instances removed.In this list 10 → 10 → 20 → 30 → 10, performing removeAll(10)will remove all three 10's and the list will look like this: 20 → 30.Then, it will return the number of instances removed, in thiscase, 3.• int contains(int num)This will return the position of the first instance of the elementnum in the list. If num is not found, return 0. In the example,having the method contains(30) will return 2 as it is located inthe second position.

The given problem is related to C++ progamming that has the methods of linked list implemented in…

Answered: plain the flow of the code. See…

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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Related questions

Question

Explain the flow of the code. See attached photo for the problem. (This is not graded)

int removeAt(int pos) {

if (pos <= index){

int elem = 0;

if (pos == 1){

return removeHead();

}

node* currnode = head;

node* prevnode = NULL;

int count = 0;

while (currnode != NULL) {

count++;

if (count == pos) {

elem = currnode-> element;

prevnode->next = currnode->next;

if (currnode == tail) {

tail = prevnode;

}

free(currnode);

index--;

return elem;

}else {

prevnode = currnode;

currnode = currnode->next;

}

return -1;

}

int removeAll(int num) {

int count=0;

node* currnode = head;

node* prevnode = NULL;

while (currnode != NULL) {

if (currnode->element == num) {

count++;

if (prevnode != NULL) {

if (currnode == tail) {

tail = prevnode;

}

node* rem = currnode;

prevnode->next = currnode->next;

currnode = currnode->next;

free(rem);

index--;

} else {

node* rem = currnode;

currnode = rem->next;

removeHead();

}

} else {

prevnode = currnode;

currnode = currnode->next;

}

return count;

}

int contains(int num) {

for (int i=1; i <= index; i++){

if (get(i) == num){

return i;

}

return 0;

}

int count(int num) {

int count =0;

for (int i =1; i <= index; i++){

if (get(i) == num){

count++;

}

return count;

}

bool move(int pos1, int pos2) {

if ( pos1 <= index && pos2 <= index){

int elem = removeAt(pos1);

if(pos2 == 1){

addHead(elem);

}

else{

addAt(elem,pos2);

}

return true;

}

return false;

}

int removeTail() {

int elem=0;

node* currnode = head;

node* prevnode = NULL;

while (currnode != NULL) {

if (currnode == tail) {

elem = currnode-> element;

if (prevnode != NULL){

prevnode->next = currnode->next;

}

tail = prevnode;

if(tail == NULL){

head = NULL;

}

free(currnode);

index--;

return elem;

}else {

prevnode = currnode;

currnode = currnode->next;

}

return 0;

}

bool isEmpty() {

if (index == 0){

return true;

}

return false;

}

void clear() {

while(head != NULL){

node* temp = head->next;

removeHead();

head = temp;

}

int count(int num)
This will return the count of the instances of the element num
in the list. In the linked list in removeAll method, having the
method count(10) will return 3 as there are three 10's in the
linked list.
Additionally, you are to implement these methods:
bool move(int pos1, int pos2)
This method will move the element at position pos1 to the
position specified as pos2 and will return true if the move is
successful. This will return false if either the specified
positions is invalid. In the example stated, operating the
method move(1, 3) will move the 1st element and shall now
become the 3rd element. The linked list shall now look like: 30
→ 40 + 10 + 50
• int removeTail()
This method will remove the last element in your linked list and
return the said element. If the list is empty, return 0. You can
use the existing methods.
• void isEmpty)
This method will return true if the linked list is empty,
otherwise return false.
• void clear()
This method will empty your linked list. Effectively, this should
and already has been called in your destructor (ie., the
-LinkedList() method) so that it will deallocate the nodes
created first before deallocating the linked list itself.

• int removeAt(int pos)
Removes the number in the posth position of the list and
returns the element removed.
Performing removeAt(3) in the example list will remove the 3rd
element of the linked list and the updated list will be: 10 → 30 →
50
When the value of pos is greater than the size or less than one,
return -1.
• int removeAll(int num)
Removes all instances of num in the linked list and returns the
number of instances removed.
In this list 10 → 10 → 20 → 30 → 10, performing removeAll(10)
will remove all three 10's and the list will look like this: 20 → 30.
Then, it will return the number of instances removed, in this
case, 3.
• int contains(int num)
This will return the position of the first instance of the element
num in the list. If num is not found, return 0. In the example,
having the method contains(30) will return 2 as it is located in
the second position.

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