C++Write a function void headEnqueue(Queue q, int key) which enqu.pdf
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C++ Write a function void headEnqueue(Queue *q, int key) which enqueues the key such that, at function exit, it is positioned at the head of the queue rather than the tail; and the order and integrity of the queue is otherwise maintained. Solution struct node { int data; node* next; }; class Queue //linked_list class { private: node* front; node* rear; public: Queue(); void insert(int); headEnqueue(Queue *q, int key); }; Queue::Queue() { front = rear = NULL; } void Queue::insert(int val) { node* temp = new node; temp->data = val; temp->next = NULL; if (front == NULL) front= rear = temp; else { node* temp1 = front; while (temp1->next) temp1 = temp1->next; temp1->next = temp; } } void Queue::headEnqueue(Queue *q, int key) { node* temp = new node; temp->data = key; temp->next=q->front; q->front=temp; }.
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#includeiostream struct node { char value; struct no.pdf
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
Implement of c & its coding programming by sarmad baloch
Implement of c & its coding programming by sarmad baloch
#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
solution in c++program Program to implement a queue using two .pdf
solution in c++
program
/* Program to implement a queue using two stacks */
#include
#include
using namespace std;
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack is empty \ \");
getchar();
exit(0);
}
else{
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
for(int i=0;i<15;i++)
{
enQueue(q, i);
}
//enQueue(q, 2);
//enQueue(q, 3);
//enQueue(q,4);
/* Dequeue items */
printf(\"the dequeue items\");
for(int i=0;i<16;i++)
{
printf(\"%d \", deQueue(q));
//printf(\"%d \", deQueue(q));
}
getchar();
}
output
the dequeue items
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
q is empty
Solution
solution in c++
program
/* Program to implement a queue using two stacks */
#include
#include
using namespace std;
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q .
Double linked list header file below for FYI#include iostream.pdf
Double linked list header file below for FYI
#include
#include
#include
#include
#ifndef DOUBLYSKIPLIST_H
#define DOUBLYSKIPLIST_H
using namespace std;
class Node {
public:
int value;
int level;
vector next;
vector prev;
Node(int val, int lvl) : value(val), level(lvl) {
next.resize(lvl + 1, nullptr);
prev.resize(lvl + 1, nullptr);
}
};
class SkipList {
public:
int level;
Node* head;
Node* tail;
int MAXIMUM_ALLOWED_LEVELS;
SkipList(int maxLevels) : MAXIMUM_ALLOWED_LEVELS(maxLevels), level(0) {
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
tail = new Node(INT_MAX, MAXIMUM_ALLOWED_LEVELS);
for (int i = 0; i <= MAXIMUM_ALLOWED_LEVELS; i++) {
head->next[i] = tail;
tail->prev[i] = head;
}
}
int RandomLevel() {
float probability = (float)rand() / RAND_MAX;
int lvl = 0;
while (probability < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS) {
lvl++;
probability = (float)rand() / RAND_MAX;
}
return lvl;
}
Node* CreateNode(int value, int level) {
return new Node(value, level);
}
void InsertElement(int value) {
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
if (current->value != value) {
int ranLevel = RandomLevel();
if (ranLevel > level) {
for (int i = level + 1; i <= ranLevel; i++) {
update[i] = head;
}
level = ranLevel;
}
Node* n = CreateNode(value, ranLevel);
for (int i = 0; i <= ranLevel; i++) {
n->next[i] = update[i]->next[i];
n->prev[i] = update[i];
update[i]->next[i]->prev[i] = n;
update[i]->next[i] = n;
}
}
}
void Delete(int value) {
if (!Search(value)) {
cout << value << " does not exist" << endl;
return;
}
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
for (int i = 0; i <= level; i++) {
if (update[i]->next[i] == current) {
update[i]->next[i] = current->next[i];
current->next[i]->prev[i] = update[i];
}
}
delete current;
}
bool Search(int value) {
Node* current = head;
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
}
current = current->next[0];
return current->value == value;
}
void Show() {
for (int i = 0; i <= level; i++) {
Node* node = head->next[i];
cout << "Level " << i << ": ";
while (node != tail) {
cout << node->value << " -> ";
node = node->next[i];
}
cout << "Tail" << endl;
}
}
void ShowBackwards() {
Node* currTail = tail;
for (int i = 0; i <= level; i++) {
Node* node = currTail->prev[i];
cout << "Level " << i << ": ";
while (node != head) {
cout << node->value << " <- ";
node = node->prev[i];
}
cout << "Head" << endl;
}
}
};
#endif
PriorityQueue header file code below
#ifndef PRIORITYQ_H
#define PRIORITYQ_H
#include
#include
#include // for INT_MAX
#include // for rand()
#include
using namespace std;
class Node {
public:
.
Recommended
#includeiostream struct node { char value; struct no.pdf
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2);
p1->insertFront(\'a\');
p1->print();
system(\"pause\");
return 0;
}
Solution
#include
struct node
{
char value;
struct node *next;
};
class StringOfNode
{
struct node * head;
public:
StringOfNode()
{
head=NULL;
}
int size()
{
struct node *p=head;
int count=1;
while(p)
{
count++;
p=p->next;
}
return count;
}
struct node * begin()
{
return head;
}
char &operator[](int index)
{
int i=0;
struct node *p=head;
while(index!=i)
{
p=p->next;
i++;
}
return p->value;
}
void insertFront(char valuee)
{
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(head==NULL)
head=temp;
else
{
temp->next=head;
head=temp;
}
}
void insert(char valuee, int pos)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
int i=0;
if(p==NULL)
{
head=temp;
}
else
{
while(i!=pos-1)
{
p=p->next;
i++;
}
temp->next=p->next;
p->next=temp;
}
}
void insertBack(char valuee)
{
struct node *p=head;
struct node * temp=(struct node *)malloc(sizeof(struct node));
temp->value=valuee;
temp->next=NULL;
if(p==NULL)
{
head=temp;
}
else
{
while(p->next!=NULL)
p=p->next;
p->next=temp;
}
}
void print()
{
struct node *p=head;
while(p)
{
std::cout<value<<\"\ \";
p=p->next;
}
}
StringOfNode operator+(const StringOfNode& b)
{
StringOfNode *S;
struct node *p=head;
while(p->next!=NULL)
p=p->next;
p->next=S->begin();
}
};
int main()
{
StringOfNode *p1=new StringOfNode;
p1->insertBack(\'b\');
p1->insertBack(\'c\');
p1->insertBack(\'e\');
p1->insert(\'d\',2.
Implement of c & its coding programming by sarmad baloch
Implement of c & its coding programming by sarmad baloch
#includeiostream #includecstdio #includecstdlib using na.pdf
#include
#include
#include
using namespace std;
struct node
{
int val;
struct node *next;
}*startNode;
// declaring class
class LinkedList
{
public:
node* create_node(int);
void insert_last();
void sort();
void search();
void reverse();
void display();
LinkedList()
{
startNode = NULL;
}
};
// main
main()
{
int userChoice, linkedListNodes, i;
LinkedList LinkList;
startNode = NULL;
while (1)
{
// displaying menu
cout<<\" 1)Insert\";
cout<<\"2) Remove\";
cout<<\"3) Search\";
cout<<\"4) Display\";
cout<<\"5) Reverse\";
cout<<\"6) Exit \";
// reading user choice
cout<<\"Enter your userChoice : \";
cin>>userChoice;
switch(userChoice)
{
case 1:
cout<<\"Inserting node to linked list: \";
LinkList.insert_begin();
cout;
break;
case 2:
cout<<\"Delete element from linked list: \";
LinkList.delete();
break;
case 3:
cout<<\"Searching data in linked list: \";
LinkList.search();
cout;
break;
case 4:
cout<<\"Displaying linked list data: \";
LinkList.display();
cout;
break;
case 5:
cout<<\"Linked list reversing: \";
LinkList.reverse();
cout;
break;
case 6:
cout<<\"Exiting...\";
exit(1);
break;
default:
cout<<\"Enter correct userChoice\";
}
}
}
// node creating to store into list
node *LinkedList::create_node(int val)
{
struct node *tempNode, *tempNode;
tempNode = new(struct node);
if (tempNode == NULL)
{
cout<<\"No memory allocated \";
return 0;
}
else
{
tempNode->val = val;
tempNode->next = NULL;
return tempNode;
}
}
/*
* Inserting Node at last
*/
void LinkedList::insert_last()
{
int val;
cout<<\"Enter the val to be inserted: \";
cin>>val;
struct node *tempNode, *tempNode;
tempNode = create_node(val);
tempNode = startNode;
while (tempNode->next != NULL)
{
tempNode = tempNode->next;
}
tempNode->next = NULL;
tempNode->next = tempNode;
cout<<\"Element Inserted\";
}
// deleting element
void LinkedList::delete()
{
int position, i, count = 0;
if (startNode == NULL)
{
cout<<\"Linkedlist empty\"<>position;
struct node *tempNode, *ptr;
tempNode = startNode;
if (position == 1)
{
startNode = tempNode->next;
}
else
{
while (tempNode != NULL)
{
tempNode = tempNode->next;
count++;
}
if (position > 0 && position <= count)
{
tempNode = startNode;
for (i = 1;i < position;i++)
{
ptr = tempNode;
tempNode = tempNode->next;
}
ptr->next = tempNode->next;
}
else
{
cout<<\"Enter correct position\";
}
}
}
// search data in linked list
void LinkedList::search()
{
int val, position = 0;
bool flagValue = false;
if (startNode == NULL)
{
cout<<\"List is empty\";
return;
}
cout<<\"Enter the val to search: \";
cin>>val;
struct node *tempNode;
tempNode = startNode;
while (tempNode != NULL)
{
position++;
if (tempNode->val == val)
{
flagValue = true;
cout<<\"Data \"<next;
}
if (!flagValue)
cout<<\"data not found\";
}
// linkedlist reverese
void LinkedList::reverse()
{
struct node *pointer1, *pointer2, *pointer3;
if (startNode == NULL)
{
cout<<\"List is empty\"<next == NULL)
{
return;
}
pointer1 = startNode;
pointer2 = pointer1->next;
pointer3 = pointer2->next;
pointer1->next = NULL;
pointer2->.
solution in c++program Program to implement a queue using two .pdf
solution in c++
program
/* Program to implement a queue using two stacks */
#include
#include
using namespace std;
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack is empty \ \");
getchar();
exit(0);
}
else{
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
for(int i=0;i<15;i++)
{
enQueue(q, i);
}
//enQueue(q, 2);
//enQueue(q, 3);
//enQueue(q,4);
/* Dequeue items */
printf(\"the dequeue items\");
for(int i=0;i<16;i++)
{
printf(\"%d \", deQueue(q));
//printf(\"%d \", deQueue(q));
}
getchar();
}
output
the dequeue items
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
q is empty
Solution
solution in c++
program
/* Program to implement a queue using two stacks */
#include
#include
using namespace std;
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q .
Double linked list header file below for FYI#include iostream.pdf
Double linked list header file below for FYI
#include
#include
#include
#include
#ifndef DOUBLYSKIPLIST_H
#define DOUBLYSKIPLIST_H
using namespace std;
class Node {
public:
int value;
int level;
vector next;
vector prev;
Node(int val, int lvl) : value(val), level(lvl) {
next.resize(lvl + 1, nullptr);
prev.resize(lvl + 1, nullptr);
}
};
class SkipList {
public:
int level;
Node* head;
Node* tail;
int MAXIMUM_ALLOWED_LEVELS;
SkipList(int maxLevels) : MAXIMUM_ALLOWED_LEVELS(maxLevels), level(0) {
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
tail = new Node(INT_MAX, MAXIMUM_ALLOWED_LEVELS);
for (int i = 0; i <= MAXIMUM_ALLOWED_LEVELS; i++) {
head->next[i] = tail;
tail->prev[i] = head;
}
}
int RandomLevel() {
float probability = (float)rand() / RAND_MAX;
int lvl = 0;
while (probability < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS) {
lvl++;
probability = (float)rand() / RAND_MAX;
}
return lvl;
}
Node* CreateNode(int value, int level) {
return new Node(value, level);
}
void InsertElement(int value) {
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
if (current->value != value) {
int ranLevel = RandomLevel();
if (ranLevel > level) {
for (int i = level + 1; i <= ranLevel; i++) {
update[i] = head;
}
level = ranLevel;
}
Node* n = CreateNode(value, ranLevel);
for (int i = 0; i <= ranLevel; i++) {
n->next[i] = update[i]->next[i];
n->prev[i] = update[i];
update[i]->next[i]->prev[i] = n;
update[i]->next[i] = n;
}
}
}
void Delete(int value) {
if (!Search(value)) {
cout << value << " does not exist" << endl;
return;
}
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
for (int i = 0; i <= level; i++) {
if (update[i]->next[i] == current) {
update[i]->next[i] = current->next[i];
current->next[i]->prev[i] = update[i];
}
}
delete current;
}
bool Search(int value) {
Node* current = head;
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
}
current = current->next[0];
return current->value == value;
}
void Show() {
for (int i = 0; i <= level; i++) {
Node* node = head->next[i];
cout << "Level " << i << ": ";
while (node != tail) {
cout << node->value << " -> ";
node = node->next[i];
}
cout << "Tail" << endl;
}
}
void ShowBackwards() {
Node* currTail = tail;
for (int i = 0; i <= level; i++) {
Node* node = currTail->prev[i];
cout << "Level " << i << ": ";
while (node != head) {
cout << node->value << " <- ";
node = node->prev[i];
}
cout << "Head" << endl;
}
}
};
#endif
PriorityQueue header file code below
#ifndef PRIORITYQ_H
#define PRIORITYQ_H
#include
#include
#include // for INT_MAX
#include // for rand()
#include
using namespace std;
class Node {
public:
.
reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx
reverse the linked list (2,4,8,10) by:
stack;
iteration;
recursion.
Using C++
Solution
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#includ<stack>
using namespace std;
struct Node {
int value;
struct Node* next;
}node;
struct Node* reverse(struct Node* head) {Â Â // Iteration Method
struct Node* current,*prev;
current = head;
prev = NULL;
while(current!=NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next; }
head = prev;-
return head; }
struct Node* insert(Node* head,int value) {
Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->value = value;
temp->next = NULL;
if(head = = NULL)
head = temp;
else {
Node*temp1=head;
while(temp1->next!=NULL)
temp1 = temp1->next;
temp1->next = temp;
} return head;
}
public void print(Node* head) {
while(head!=NULL) {
cout<<head->value;
head = head->next; }
}
struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method
if(head==NULL || head->next ==NULL) return;
Node*first = head;
Node *rest = head->next;
if(rest !=NULL) return;
recursiveReverse(rest);
first ->next->next = first;
first->next =NULL;
head = rest;
return head; }
void Reverse(struct Node* head)Â Â // stack Implementation
{
stack<node*> Stack;
node *traverse = head;
while(traverse!=NULL) {
Stack.push(traverse);
traverse = traverse->next;Â Â }
node *temp = Stack.top();
head = temp;
Stack.pop();
while(!Stack.empty()) {
temp->next = Stack.top();
Stack.pop();
temp = temp->next;Â Â }
temp->next = NULL;
return Stack; }
int main() {
struct Node* head =NULL;
head = insert(head,2);
head = insert(head,4);
head = insert(head,8);
head = insert(head,10);
print(head);
head= reverse(head);
print(head);
head = recursiveReverse(head);
print(head);
head = void Reverse(head);
void print(head);
return 0;
}
.
C++ Program to Implement Singly Linked List #includei.pdf
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<<\"1.Insert Node at beginning\"<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (ptr->info > s->info)
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
/*
* Delete element at a given position
*/
void single_llist::delete_pos()
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<>pos;
cout<<\"Enter the new value: \";
cin>>value;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start->info = value;
}
else
{
for (i = 0;i < pos - 1;i++)
{
if (s == NULL)
{
cout<<\"There are less than \"<next;
}
s->info = value;
}
cout<<\"Node Updated\"<>value;
struct node *s;
s = start;
while (s != NULL)
{
pos++;
if (s->info == value)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2 = ptr3;
ptr3 = ptr3->next;
ptr2->next = ptr1;
}
start = ptr2;
}
/*
* Display Elements of a link list
*/
vo.
#include iostream #includestdlib.h using namespace std;str.pdf
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node *head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
if(prev==NULL){
head2 = curr;
globalhead = curr;
}else{
prev->next = curr;
}
prev = curr;
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<
Solution
//Tested on Windos OS Dev c++
/*********************LinkedList.cpp********************/
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node **head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
curr->next = NULL;
if(*head2==NULL){
*head2=curr;
}else{
prev=*head2;
while(prev->next!=NULL){
prev=prev->next;
}
prev->next=curr;
}
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<.
double linked list header file code#include iostream#include.pdf
double linked list header file code
#include
#include
#include
#include
#ifndef DOUBLYSKIPLIST_H
#define DOUBLYSKIPLIST_H
using namespace std;
class Node {
public:
int value;
int level;
vector next;
vector prev;
Node(int val, int lvl) : value(val), level(lvl) {
next.resize(lvl + 1, nullptr);
prev.resize(lvl + 1, nullptr);
}
};
class SkipList {
public:
int level;
Node* head;
Node* tail;
int MAXIMUM_ALLOWED_LEVELS;
SkipList(int maxLevels) : MAXIMUM_ALLOWED_LEVELS(maxLevels), level(0) {
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
tail = new Node(INT_MAX, MAXIMUM_ALLOWED_LEVELS);
for (int i = 0; i <= MAXIMUM_ALLOWED_LEVELS; i++) {
head->next[i] = tail;
tail->prev[i] = head;
}
}
int RandomLevel() {
float probability = (float)rand() / RAND_MAX;
int lvl = 0;
while (probability < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS) {
lvl++;
probability = (float)rand() / RAND_MAX;
}
return lvl;
}
Node* CreateNode(int value, int level) {
return new Node(value, level);
}
void InsertElement(int value) {
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
if (current->value != value) {
int ranLevel = RandomLevel();
if (ranLevel > level) {
for (int i = level + 1; i <= ranLevel; i++) {
update[i] = head;
}
level = ranLevel;
}
Node* n = CreateNode(value, ranLevel);
for (int i = 0; i <= ranLevel; i++) {
n->next[i] = update[i]->next[i];
n->prev[i] = update[i];
update[i]->next[i]->prev[i] = n;
update[i]->next[i] = n;
}
}
}
void Delete(int value) {
if (!Search(value)) {
cout << value << " does not exist" << endl;
return;
}
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
for (int i = 0; i <= level; i++) {
if (update[i]->next[i] == current) {
update[i]->next[i] = current->next[i];
current->next[i]->prev[i] = update[i];
}
}
delete current;
}
bool Search(int value) {
Node* current = head;
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
}
current = current->next[0];
return current->value == value;
}
void Show() {
for (int i = 0; i <= level; i++) {
Node* node = head->next[i];
cout << "Level " << i << ": ";
while (node != tail) {
cout << node->value << " -> ";
node = node->next[i];
}
cout << "Tail" << endl;
}
}
void ShowBackwards() {
Node* currTail = tail;
for (int i = 0; i <= level; i++) {
Node* node = currTail->prev[i];
cout << "Level " << i << ": ";
while (node != head) {
cout << node->value << " <- ";
node = node->prev[i];
}
cout << "Head" << endl;
}
}
};
#endif
Priority Queue header file code
#ifndef PRIORITYQ_H
#define PRIORITYQ_H
#include
#include
#include // for INT_MAX
#include // for rand()
#include
#include "doublySkiplist.h"
using namespace std;
class N.
package algs13;import stdlib.;import java.util.Iterator;im.docx
package algs13;
import stdlib.*;
import java.util.Iterator;
import java.util.NoSuchElementException;
/* ***********************************************************************
* Compilation: javac Queue.java
* Execution: java Queue < input.txt
* Data files: http://algs4.cs.princeton.edu/13stacks/tobe.txt
*
* A generic queue, implemented using a linked list.
*
* % java Queue < tobe.txt
* to be or not to be (2 left on queue)
*
*************************************************************************/
/**
* The <tt>Queue</tt> class represents a first-in-first-out (FIFO)
* queue of generic items.
* It supports the usual <em>enqueue</em> and <em>dequeue</em>
* operations, along with methods for peeking at the top item,
* testing if the queue is empty, and iterating through
* the items in FIFO order.
* <p>
* All queue operations except iteration are constant time.
* <p>
* For additional documentation, see <a href="http://algs4.cs.princeton.edu/13stacks">Section 1.3</a> of
* <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
*/
public class Queue<T> implements Iterable<T> {
private int N; // number of elements on queue
private Node<T> first; // beginning of queue
private Node<T> last; // end of queue
// helper linked list class
private static class Node<T> {
public Node() { }
public T item;
public Node<T> next;
}
/**
* Create an empty queue.
*/
public Queue() {
first = null;
last = null;
N = 0;
}
/**
* Is the queue empty?
*/
public boolean isEmpty() {
return first == null;
}
/**
* Return the number of items in the queue.
*/
public int size() {
return N;
}
/**
* Return the item least recently added to the queue.
* @throws java.util.NoSuchElementException if queue is empty.
*/
public T peek() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
return first.item;
}
/**
* Add the item to the queue.
*/
public void enqueue(T item) {
Node<T> oldlast = last;
last = new Node<>();
last.item = item;
last.next = null;
if (isEmpty()) first = last;
else oldlast.next = last;
N++;
}
/**
* Remove and return the item on the queue least recently added.
* @throws java.util.NoSuchElementException if queue is empty.
*/
public T dequeue() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
T item = first.item;
first = first.next;
N--;
if (isEmpty()) last = null;
return item;
}
/**
* Return string representation.
*/
public String toString() {
StringBuilder s = new StringBuilder();
for (T item : this)
s.append(item + " ");
return s.toString();
}
// check internal invariants
private static <T> boolean check(Queue<T> that) {
int N = that.N;
Queue.Node<T> first = that.first;
Queue.Node<T> last = that.last;
if (N == 0) {
if (first != null) return false;
if (last != null) return false;
}
else if (N == 1) {
if (f.
I l Show that if X has the discrete topology, then its only conn.pdf
I l
Show that if X has the discrete topology, then its only connected subspaces are point sets.
Solution
Let X have discrete topology.
Any subset Y of X is the union of singletons , each of which is both open and closed (as X has
discrete topology)
So the maximal sets which cannot expressed as a disjoint union of open (closed) subsets are the
singletons.
Hence the connected components of X are the singletons..
In 2006 the CEO of Bear Sterns, James Caynes, received a compensatio.pdf
In 2006 the CEO of Bear Sterns, James Caynes, received a compensation package of $34
million. The following year Bear Sterns cost $2.7 billion to the taxpayers. In 2006, the CEO of
Lehman Brothers received a compensation package of $27 million. On September 15, 2008,
Lehman Brothers filed for bankruptcy. The collapse of Lehman Brothers is seen by many as the
key event that sparked the Global Financial Crisis. In 2006, the CEO of Citigroup, Charles
Prince, received a compensation package of $25 million. Since then the stock price has fallen
from $50 a share to $3.5 a share. The CEO of Countrywide Financial, Angelo Mozilo, did even
better. His compensation package was $43 million. Angelo Mozilo and two other top executives
were charged by the Security and Exchange Commission (SEC) with fraud. According to the
SEC, from 2005 through 2007, Countrywide Financial engaged in an unprecedented expansion
of its underwriting guidelines and was writing riskier and riskier loans, which these senior
executives were warned might ultimately curtail the company\'s ability to sell them. Countrywide
Financial was the third biggest originator of subprime mortgages and the nation\'s leader in
subprime mortgage- backed securities. The tragedy is that these individuals did not make
decisions that were in their companies’ best interest. Why? What went wrong? What caused the
relation between the CEO and the stockholders to go so badly awry? Discuss
Solution
Leaders who lose their way are not necessarily bad people; rather, they lose their moral bearings,
often yielding to seductions in their paths. Very few people go into leadership roles to cheat or
do evil, yet we all have the capacity for actions we deeply regret unless we stay grounded.
Leaders whose goal is the quest for power over others, unlimited wealth, or the fame that comes
with success tend to look to others to gain satisfaction, and often appear self-centered and
egotistical. They start to believe their own press. As leaders of institutions, they eventually
believe the institution cannot succeed without them.
While most people value fair compensation for their accomplishments, few leaders start out
seeking only money, power, and prestige. Along the way, the rewards—bonus checks,
newspaper articles, perks, and stock appreciation—fuel increasing desires for more.
This creates a deep desire to keep it going, often driven by desires to overcome narcissistic
wounds from childhood. Many times, this desire is so strong that leaders breach the ethical
standards that previously governed their conduct, which can be bizarre and even illegal.
Many leaders get to the top by imposing their will on others, even destroying people standing in
their way. When they reach the top, they may be paranoid that others are trying to knock them
off their pedestal. Sometimes they develop an impostor complex, caused by deep insecurities that
they aren\'t good enough and may be unmasked.
To prove they aren\'t impostors, they drive so.
More Related Content
Similar to C++Write a function void headEnqueue(Queue q, int key) which enqu.pdf
reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx
reverse the linked list (2,4,8,10) by:
stack;
iteration;
recursion.
Using C++
Solution
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#includ<stack>
using namespace std;
struct Node {
int value;
struct Node* next;
}node;
struct Node* reverse(struct Node* head) {Â Â // Iteration Method
struct Node* current,*prev;
current = head;
prev = NULL;
while(current!=NULL) {
next = current->next;
current->next = prev;
prev = current;
current = next; }
head = prev;-
return head; }
struct Node* insert(Node* head,int value) {
Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->value = value;
temp->next = NULL;
if(head = = NULL)
head = temp;
else {
Node*temp1=head;
while(temp1->next!=NULL)
temp1 = temp1->next;
temp1->next = temp;
} return head;
}
public void print(Node* head) {
while(head!=NULL) {
cout<<head->value;
head = head->next; }
}
struct Node*recursiveReverse(struct Node* &head) {Â Â //Recursion method
if(head==NULL || head->next ==NULL) return;
Node*first = head;
Node *rest = head->next;
if(rest !=NULL) return;
recursiveReverse(rest);
first ->next->next = first;
first->next =NULL;
head = rest;
return head; }
void Reverse(struct Node* head)Â Â // stack Implementation
{
stack<node*> Stack;
node *traverse = head;
while(traverse!=NULL) {
Stack.push(traverse);
traverse = traverse->next;Â Â }
node *temp = Stack.top();
head = temp;
Stack.pop();
while(!Stack.empty()) {
temp->next = Stack.top();
Stack.pop();
temp = temp->next;Â Â }
temp->next = NULL;
return Stack; }
int main() {
struct Node* head =NULL;
head = insert(head,2);
head = insert(head,4);
head = insert(head,8);
head = insert(head,10);
print(head);
head= reverse(head);
print(head);
head = recursiveReverse(head);
print(head);
head = void Reverse(head);
void print(head);
return 0;
}
.
C++ Program to Implement Singly Linked List #includei.pdf
/*
* C++ Program to Implement Singly Linked List
*/
#include
#include
#include
using namespace std;
/*
* Node Declaration
*/
struct node
{
int info;
struct node *next;
}*start;
/*
* Class Declaration
*/
class single_llist
{
public:
node* create_node(int);
void insert_begin();
void insert_pos();
void insert_last();
void delete_pos();
void sort();
void search();
void update();
void reverse();
void display();
single_llist()
{
start = NULL;
}
};
/*
* Main :contains menu
*/
main()
{
int choice, nodes, element, position, i;
single_llist sl;
start = NULL;
while (1)
{
cout<<\"1.Insert Node at beginning\"<>choice;
switch(choice)
{
case 1:
cout<<\"Inserting Node at Beginning: \"<info = value;
temp->next = NULL;
return temp;
}
}
/*
* Inserting element in beginning
*/
void single_llist::insert_begin()
{
int value;
cout<<\"Enter the value to be inserted: \";
cin>>value;
struct node *temp, *p;
temp = create_node(value);
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
p = start;
start = temp;
start->next = p;
}
cout<<\"Element Inserted at beginning\"<>value;
struct node *temp, *s;
temp = create_node(value);
s = start;
while (s->next != NULL)
{
s = s->next;
}
temp->next = NULL;
s->next = temp;
cout<<\"Element Inserted at last\"<>value;
struct node *temp, *s, *ptr;
temp = create_node(value);
cout<<\"Enter the postion at which node to be inserted: \";
cin>>pos;
int i;
s = start;
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos == 1)
{
if (start == NULL)
{
start = temp;
start->next = NULL;
}
else
{
ptr = start;
start = temp;
start->next = ptr;
}
}
else if (pos > 1 && pos <= counter)
{
s = start;
for (i = 1; i < pos; i++)
{
ptr = s;
s = s->next;
}
ptr->next = temp;
temp->next = s;
}
else
{
cout<<\"Positon out of range\"<next;s !=NULL;s = s->next)
{
if (ptr->info > s->info)
{
value = ptr->info;
ptr->info = s->info;
s->info = value;
}
}
ptr = ptr->next;
}
}
/*
* Delete element at a given position
*/
void single_llist::delete_pos()
{
int pos, i, counter = 0;
if (start == NULL)
{
cout<<\"List is empty\"<>pos;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start = s->next;
}
else
{
while (s != NULL)
{
s = s->next;
counter++;
}
if (pos > 0 && pos <= counter)
{
s = start;
for (i = 1;i < pos;i++)
{
ptr = s;
s = s->next;
}
ptr->next = s->next;
}
else
{
cout<<\"Position out of range\"<>pos;
cout<<\"Enter the new value: \";
cin>>value;
struct node *s, *ptr;
s = start;
if (pos == 1)
{
start->info = value;
}
else
{
for (i = 0;i < pos - 1;i++)
{
if (s == NULL)
{
cout<<\"There are less than \"<next;
}
s->info = value;
}
cout<<\"Node Updated\"<>value;
struct node *s;
s = start;
while (s != NULL)
{
pos++;
if (s->info == value)
{
flag = true;
cout<<\"Element \"<next;
}
if (!flag)
cout<<\"Element \"<next == NULL)
{
return;
}
ptr1 = start;
ptr2 = ptr1->next;
ptr3 = ptr2->next;
ptr1->next = NULL;
ptr2->next = ptr1;
while (ptr3 != NULL)
{
ptr1 = ptr2;
ptr2 = ptr3;
ptr3 = ptr3->next;
ptr2->next = ptr1;
}
start = ptr2;
}
/*
* Display Elements of a link list
*/
vo.
#include iostream #includestdlib.h using namespace std;str.pdf
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node *head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
if(prev==NULL){
head2 = curr;
globalhead = curr;
}else{
prev->next = curr;
}
prev = curr;
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<
Solution
//Tested on Windos OS Dev c++
/*********************LinkedList.cpp********************/
#include
#include
using namespace std;
struct node { //Task 1
char val;
node *next;
};
void printList(node *head) { //Task 2
if (head == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head;
int count = 0;
while(temp != NULL) {
cout << \"Node \" << count << \": \" << temp ->val << endl;
count++;
temp = temp->next;
}
}
struct node *globalhead = NULL;
void deepCopy(node *head1, node **head2) { //Task 3
if (head1 == NULL) {
cout << \"Empty list\" << endl;
return;
}
node *temp = head1;
node *curr = NULL;
node *prev = NULL;
while(temp != NULL) {
curr = (struct node *)malloc(sizeof(struct node));
curr->val = temp->val;
curr->next = NULL;
if(*head2==NULL){
*head2=curr;
}else{
prev=*head2;
while(prev->next!=NULL){
prev=prev->next;
}
prev->next=curr;
}
temp = temp->next;
}
}
int main() { //Task 4
// Example #1
node p00, p01, p02, p03, p10, p11, p12, p13;
// List 1: A B B B
p00.val = \'A\'; p00.next = &p01;
p01.val = \'B\'; p01.next = &p02;
p02.val = \'B\'; p02.next = &p03;
p03.val = \'B\'; p03.next = NULL;
p10.next = NULL;
cout<<\"First List:\"<.
double linked list header file code#include iostream#include.pdf
double linked list header file code
#include
#include
#include
#include
#ifndef DOUBLYSKIPLIST_H
#define DOUBLYSKIPLIST_H
using namespace std;
class Node {
public:
int value;
int level;
vector next;
vector prev;
Node(int val, int lvl) : value(val), level(lvl) {
next.resize(lvl + 1, nullptr);
prev.resize(lvl + 1, nullptr);
}
};
class SkipList {
public:
int level;
Node* head;
Node* tail;
int MAXIMUM_ALLOWED_LEVELS;
SkipList(int maxLevels) : MAXIMUM_ALLOWED_LEVELS(maxLevels), level(0) {
head = new Node(INT_MIN, MAXIMUM_ALLOWED_LEVELS);
tail = new Node(INT_MAX, MAXIMUM_ALLOWED_LEVELS);
for (int i = 0; i <= MAXIMUM_ALLOWED_LEVELS; i++) {
head->next[i] = tail;
tail->prev[i] = head;
}
}
int RandomLevel() {
float probability = (float)rand() / RAND_MAX;
int lvl = 0;
while (probability < 0.5 && lvl < MAXIMUM_ALLOWED_LEVELS) {
lvl++;
probability = (float)rand() / RAND_MAX;
}
return lvl;
}
Node* CreateNode(int value, int level) {
return new Node(value, level);
}
void InsertElement(int value) {
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
if (current->value != value) {
int ranLevel = RandomLevel();
if (ranLevel > level) {
for (int i = level + 1; i <= ranLevel; i++) {
update[i] = head;
}
level = ranLevel;
}
Node* n = CreateNode(value, ranLevel);
for (int i = 0; i <= ranLevel; i++) {
n->next[i] = update[i]->next[i];
n->prev[i] = update[i];
update[i]->next[i]->prev[i] = n;
update[i]->next[i] = n;
}
}
}
void Delete(int value) {
if (!Search(value)) {
cout << value << " does not exist" << endl;
return;
}
Node* current = head;
vector update(MAXIMUM_ALLOWED_LEVELS + 1, nullptr);
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
update[i] = current;
}
current = current->next[0];
for (int i = 0; i <= level; i++) {
if (update[i]->next[i] == current) {
update[i]->next[i] = current->next[i];
current->next[i]->prev[i] = update[i];
}
}
delete current;
}
bool Search(int value) {
Node* current = head;
for (int i = level; i >= 0; i--) {
while (current->next[i]->value < value) {
current = current->next[i];
}
}
current = current->next[0];
return current->value == value;
}
void Show() {
for (int i = 0; i <= level; i++) {
Node* node = head->next[i];
cout << "Level " << i << ": ";
while (node != tail) {
cout << node->value << " -> ";
node = node->next[i];
}
cout << "Tail" << endl;
}
}
void ShowBackwards() {
Node* currTail = tail;
for (int i = 0; i <= level; i++) {
Node* node = currTail->prev[i];
cout << "Level " << i << ": ";
while (node != head) {
cout << node->value << " <- ";
node = node->prev[i];
}
cout << "Head" << endl;
}
}
};
#endif
Priority Queue header file code
#ifndef PRIORITYQ_H
#define PRIORITYQ_H
#include
#include
#include // for INT_MAX
#include // for rand()
#include
#include "doublySkiplist.h"
using namespace std;
class N.
package algs13;import stdlib.;import java.util.Iterator;im.docx
package algs13;
import stdlib.*;
import java.util.Iterator;
import java.util.NoSuchElementException;
/* ***********************************************************************
* Compilation: javac Queue.java
* Execution: java Queue < input.txt
* Data files: http://algs4.cs.princeton.edu/13stacks/tobe.txt
*
* A generic queue, implemented using a linked list.
*
* % java Queue < tobe.txt
* to be or not to be (2 left on queue)
*
*************************************************************************/
/**
* The <tt>Queue</tt> class represents a first-in-first-out (FIFO)
* queue of generic items.
* It supports the usual <em>enqueue</em> and <em>dequeue</em>
* operations, along with methods for peeking at the top item,
* testing if the queue is empty, and iterating through
* the items in FIFO order.
* <p>
* All queue operations except iteration are constant time.
* <p>
* For additional documentation, see <a href="http://algs4.cs.princeton.edu/13stacks">Section 1.3</a> of
* <i>Algorithms, 4th Edition</i> by Robert Sedgewick and Kevin Wayne.
*/
public class Queue<T> implements Iterable<T> {
private int N; // number of elements on queue
private Node<T> first; // beginning of queue
private Node<T> last; // end of queue
// helper linked list class
private static class Node<T> {
public Node() { }
public T item;
public Node<T> next;
}
/**
* Create an empty queue.
*/
public Queue() {
first = null;
last = null;
N = 0;
}
/**
* Is the queue empty?
*/
public boolean isEmpty() {
return first == null;
}
/**
* Return the number of items in the queue.
*/
public int size() {
return N;
}
/**
* Return the item least recently added to the queue.
* @throws java.util.NoSuchElementException if queue is empty.
*/
public T peek() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
return first.item;
}
/**
* Add the item to the queue.
*/
public void enqueue(T item) {
Node<T> oldlast = last;
last = new Node<>();
last.item = item;
last.next = null;
if (isEmpty()) first = last;
else oldlast.next = last;
N++;
}
/**
* Remove and return the item on the queue least recently added.
* @throws java.util.NoSuchElementException if queue is empty.
*/
public T dequeue() {
if (isEmpty()) throw new NoSuchElementException("Queue underflow");
T item = first.item;
first = first.next;
N--;
if (isEmpty()) last = null;
return item;
}
/**
* Return string representation.
*/
public String toString() {
StringBuilder s = new StringBuilder();
for (T item : this)
s.append(item + " ");
return s.toString();
}
// check internal invariants
private static <T> boolean check(Queue<T> that) {
int N = that.N;
Queue.Node<T> first = that.first;
Queue.Node<T> last = that.last;
if (N == 0) {
if (first != null) return false;
if (last != null) return false;
}
else if (N == 1) {
if (f.
Similar to C++Write a function void headEnqueue(Queue q, int key) which enqu.pdf (20)
reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx
reverse the linked list (2-4-8-10) by- stack- iteration- recursion- U.docx
DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY
DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY
C++ Program to Implement Singly Linked List #includei.pdf
C++ Program to Implement Singly Linked List #includei.pdf
#include iostream #includestdlib.h using namespace std;str.pdf
#include iostream #includestdlib.h using namespace std;str.pdf
double linked list header file code#include iostream#include.pdf
double linked list header file code#include iostream#include.pdf
package algs13;import stdlib.;import java.util.Iterator;im.docx
package algs13;import stdlib.;import java.util.Iterator;im.docx
More from arjuncp10
I l Show that if X has the discrete topology, then its only conn.pdf
I l
Show that if X has the discrete topology, then its only connected subspaces are point sets.
Solution
Let X have discrete topology.
Any subset Y of X is the union of singletons , each of which is both open and closed (as X has
discrete topology)
So the maximal sets which cannot expressed as a disjoint union of open (closed) subsets are the
singletons.
Hence the connected components of X are the singletons..
In 2006 the CEO of Bear Sterns, James Caynes, received a compensatio.pdf
In 2006 the CEO of Bear Sterns, James Caynes, received a compensation package of $34
million. The following year Bear Sterns cost $2.7 billion to the taxpayers. In 2006, the CEO of
Lehman Brothers received a compensation package of $27 million. On September 15, 2008,
Lehman Brothers filed for bankruptcy. The collapse of Lehman Brothers is seen by many as the
key event that sparked the Global Financial Crisis. In 2006, the CEO of Citigroup, Charles
Prince, received a compensation package of $25 million. Since then the stock price has fallen
from $50 a share to $3.5 a share. The CEO of Countrywide Financial, Angelo Mozilo, did even
better. His compensation package was $43 million. Angelo Mozilo and two other top executives
were charged by the Security and Exchange Commission (SEC) with fraud. According to the
SEC, from 2005 through 2007, Countrywide Financial engaged in an unprecedented expansion
of its underwriting guidelines and was writing riskier and riskier loans, which these senior
executives were warned might ultimately curtail the company\'s ability to sell them. Countrywide
Financial was the third biggest originator of subprime mortgages and the nation\'s leader in
subprime mortgage- backed securities. The tragedy is that these individuals did not make
decisions that were in their companies’ best interest. Why? What went wrong? What caused the
relation between the CEO and the stockholders to go so badly awry? Discuss
Solution
Leaders who lose their way are not necessarily bad people; rather, they lose their moral bearings,
often yielding to seductions in their paths. Very few people go into leadership roles to cheat or
do evil, yet we all have the capacity for actions we deeply regret unless we stay grounded.
Leaders whose goal is the quest for power over others, unlimited wealth, or the fame that comes
with success tend to look to others to gain satisfaction, and often appear self-centered and
egotistical. They start to believe their own press. As leaders of institutions, they eventually
believe the institution cannot succeed without them.
While most people value fair compensation for their accomplishments, few leaders start out
seeking only money, power, and prestige. Along the way, the rewards—bonus checks,
newspaper articles, perks, and stock appreciation—fuel increasing desires for more.
This creates a deep desire to keep it going, often driven by desires to overcome narcissistic
wounds from childhood. Many times, this desire is so strong that leaders breach the ethical
standards that previously governed their conduct, which can be bizarre and even illegal.
Many leaders get to the top by imposing their will on others, even destroying people standing in
their way. When they reach the top, they may be paranoid that others are trying to knock them
off their pedestal. Sometimes they develop an impostor complex, caused by deep insecurities that
they aren\'t good enough and may be unmasked.
To prove they aren\'t impostors, they drive so.
How would a CFE devise a plan to prevent subsequent employee fraud.pdf
How would a CFE devise a plan to prevent subsequent employee fraud?
Solution
ANSWER:
Although the prevention of fraud is always less costly and more effective than fraud detection
(and subsequent investigation), however unfortunately prevention is not always possible. This
the reason, as CFE’s Certified Fraud Examiners (CFE) we must all be heavy promoters (and
supporters) of client internal audit functions
The CFE devise a plan to internal audit that includes the procedures to identify or prevent
significant incomplete, inaccurate, or unauthorized transactions; unreliable financial reporting;
deficiencies in safeguarding assets; and deviations from regulations, laws, and institutional
policies. When effectively designed and implemented, internal audits gives directors and senior
managers with timely information in regard to weaknesses in the internal control system and thus
facilitating timely remedial action and prevent subsequent employee fraud. Each institution is
required to have an internal audit function appropriate to the size and the nature and scope of the
activities.
Hilary rode her horse for 8 miles until it was hurt.Then she walked .pdf
Hilary rode her horse for 8 miles until it was hurt.Then she walked back home to call a vet. She
figures the horse wals twice as fast as she does. the whole trip took 4 hours, so how fast would
Hilary have to walk?
Solution
The equation used in this problem is SpeedXTime=Distance
We know that the entire trip takes 4 hours
Let us designate x as the time that it takes the horse to travel 8 miles
4-x= is designated as the time it takes Hilary to walk back
we also know the ratio of rate 2:1
Let us set up the equation
(Distance/Time)=speed of the horse =8/x
(Distance/Time)=speed of Hilary =8/4-x
We need to make both sides of the eqation equal. In order to do this we need to multiply (8/4-x)
by 1/2 since the horse is going 2x the speed of Hilary
8/x=(1/2)(8/4-x)
Simplify by dividing 8 by 2 so that you get 8/x=4/4-x
Cross multiply to get 8(4-x)=4x
Use the distributive property
32-8x=4x
Add 8x to both sides and get
32=12x
Divide both sides by 12
x=1.33 hours it took the horse
4-x=2.67 hours it took Hilary
2.99 or 3 miles per hour
To double check let\'s plug in values
Speed(Time)=8
(3)2.67)=8.0 miles for Hillary
The horse speed is 8/1.33=6.0 mph
the horse is going 6/3 as fast as Hilary the answer is 2..
fully comments for my program, thank you will thumb up#include io.pdf
fully comments for my program, thank you will thumb up
#include
#include
#include
#include
using namespace std;
struct book
{
int ISBN;
string Author;
string Title;
string publisher;
int Quantity;
double price;
};
void choice1(book books[], int& size, int MAX_SIZE)
{
ifstream inFile;
inFile.open(\"inventory.txt\");
if (inFile.fail())
cout <<\"file could not open\"<> books[size].ISBN;
cout << \"Enter the author name: \";
cin >> books[size].Author;
cout << \"Enter the book tile: \";
cin >> books[size].Title;
cin.get();
cout << \"Enter the books quantity: \";
cin >> books[size].Quantity;
cout << \"Enter the book price: $\";
cin >> books[size].price;
size++;
cout << \"You have successfully inserted an entry.\" << endl;
}
}
void choice3(book books[], int& size)
{
if(size == 0)
cout << \"Array is empty. Read the data first.\" << endl;
else
{
int isbn;
cout << \"\ Enter the ISBN of the book: \";
cin >> isbn;
for(int i = 0; i < size; i++)
{
if(books[i].ISBN == isbn)
{
int j = i;
while(j < size - 1)
{
books[j] = books[j + 1];
j++;
}
size--;
break;
}
}
cout << \"You have successfully deleted an entry.\" << endl;
}
}
void choice4(book books[], int size)
{
if(size == 0)
cout << \"Array is empty. Read the data first.\" << endl;
else
{
int isbn;
int option;
int qty;
cout << \"\ Enter the ISBN of the book: \";
cin >> isbn;
cout << \"1. Increment\" << endl;
cout << \"2. Decrement\" << endl;
cout << \"3. Add New\" << endl;
cout << \"Enter your option: \";
cin >> option;
cout << \"Enter the quantity: \";
cin >> qty;
for(int i = 0; i < size; i++)
{
if(books[i].ISBN == isbn)
{
if(option == 1)
books[i].Quantity += qty;
else if(option == 2)
{
books[i].Quantity -= qty;
if(books[i].Quantity)
books[i].Quantity = 0;
}
else if(option == 3)
books[i].Quantity = qty;
break;
}
}
cout << \"You have successfully updated the array.\" << endl;
}
}
void choice5(book books[], int size)
{
for(int i = 1; i < size; i++)
{
book current = books[i];
int j = i;
while(j > 0 && (books[j - 1].Title).compare(current.Title) > 0)
{
books[j] = books[j - 1];
j--;
}
books[j] = current;
}
if(size != 0)
cout << \"You have successfully sorted the array.\" << endl;
else
cout << \"Array is empty. Read the data first.\" << endl;
}
void choice6(book books[], int size)
{
for(int i = 0; i < size; i++)
{
cout << endl;
cout << \"Book Number: \" << (i + 1) << endl;
cout << \"ISBN: \" << books[i].ISBN << endl;
cout << \"Author: \" << books[i].Author << endl;
cout << \"Title: \" << books[i].Title << endl;
cout << \"Quantity: \" << books[i].Quantity << endl;
cout << \"Price: $\" << books[i].price << endl;
}
if(size != 0)
cout << \"You have successfully printed the array.\" << endl;
else
cout << \"Array is empty. Read the file first.\" << endl;
}
void choice7(book books[], int size)
{
ofstream outFile;
outFile.open(\"finalData.dat\");
for(int i = 0; i < size; i++)
{
outFile << \"Book Number: \" << (i + 1) << endl;
outFile << \"ISBN: \" << books[i].ISBN << endl;
outFile << \"Author: \" << books[i].Author << end.
Find the admittance Yab in the circuit seen in the figure. Take that.pdf
Find the admittance Yab in the circuit seen in the figure. Take that R1 = 6 , R2 = 6 , R3 = 4 and
R4 = 12.8
Part A
Express Yab in rectangular form.
Express your answer in complex form.
The answer is NOT
Yab = (4+j3)10^(5) mS
Yab = 40+j30 mS
510^(5) e^(j36.86)Yab =
510^(5) e^(j36.86) mS j12.8 n Yab bo 12 j10 n
Solution
Z3=R3||j10=4*j10/(4+j10)
Z2=R2+j12=6+j12
Z1=R1-j2=6-j2
Zequ1=Z1||Z2||Z3=2.1517 +j 0.5793
Z=R4+Zequ1-j12.8=12.8-j12.8+2.1517 + j0.5793=14.9517 -j12.2207
Y=1/Z=1/14.9517 -j12.2207
Y=0.0401 + j0.0328
convert rect to polar
a=sqrt((0.0401^2)+(0.0328^2))=0.0518
b=tan-1(-0.0328/0.0401)=- 0.6856 in radins, if conver to degree=-39.2
ans:
0.051806*expj-39.2.
Diversity Paper Each student will complete a diversity research assig.pdf
Diversity Paper Each student will complete a diversity research assignment. Students will
choose one of the populations below and have the selection approved by the instructor. The
assignment explores the resiliency of elders of (African American antic/Latino. Asian American
Indian) or the status of older women in society compared to older men. Students will research the
economic status, health, social supports, living situations marry of one of the special populations
chosen by the student. They will also research specific cultural values and beliefs regarding life,
acceptance of help, and death. Students are encouraged to use the text and at least 3 other reliable
sources to gather information. Sources should be documented in APA style of referencing
Length is 5 pages of content, double-spaced in 12 point font. Paper is due the 7th week of class
and is worth so points. No late submissions will be accepted. The due date is Friday, February
24, by 11:00pm Central Start Date Due Date is Friday, February 24, by 11:00 pm central Time.
Solution
DIVERSITY RESEARCH ASSIGNMENT
Age and Socioeconomic Status
Socioeconomic status (SES) is often measured as a combination of education, income and
occupation. It is commonly conceptualized as the social standing or class of an individual or
group. When viewed through a social class lens, privilege, power and control are emphasized.
Furthermore, an examination of SES as a gradient or continuous variable reveals inequities in
access to and distribution of resources. SES is relevant to all realms of behavioural and social
science, including research, practice, education and advocacy.
SES Affects Our Society
SES affects overall human functioning, including our physical and mental health. Its effects can
be observed across the life span. Variance in socioeconomic status such as disparities in the
distribution of wealth, income and access to resources mitigate social problems that ultimately
affect everyone. All benefit from an increased focus on the foundations of socioeconomic
inequalities and efforts to reduce the deep gaps in socioeconomic status observed today in the
United States and abroad.
While older and retired U.S. residents may enjoy a higher standard of living than their
predecessors, older Americans remain among the most economically vulnerable groups. Of
major concern is whether older Americans will outlive their financial resources. Psychologists
and other social and behavioural science professionals possess the tools necessary to study and
identify strategies to alleviate these disparities at both individual and societal levels.
SES Impacts the Lives of Older Adults
The United States is facing unprecedented increases in the older adult population. Americans age
65 and over comprise nearly 13 percent of the U.S. population, and their proportion is estimated
to increase to 20 percent of the population in the next 25 years. As the percentage of older
Americans rises, so does concern for their econom.
Detailed solutions please 1. Let R and S be commutative rings and le.pdf
Detailed solutions please: 1. Let R and S be commutative rings and let sigma: R right arrow S be
a homomorphism. Prove that if sigma is one to one and b in R is a zero divisor of R, then
sigma(b) is a zero divisor of S.
Solution
we are given that R--->S is a homomorphism
A homomorphism is a map that preserves selected structure between two algebraic structures,
with the structure to be preserved being given by the naming of the homomorphism
the set R and S are the same as they form a homomorphic relation.
now b E R that is b lies with the set R
and since R and S are homomorphic so b will also lie in the ser S
since sigma is one to one function threfore the entire domain of sigma will be mapped .
its also gven that sigma(b) is a zero divisor of R
and since b E S as well
so sigma(b) is a zero divisor of S as well
hence proved..
Describe the mechanisms of asexual reproduction inProkaryotesPr.pdf
Describe the mechanisms of asexual reproduction in:
Prokaryotes
Protists
Plants-
Animals
Solution
Prokaryotes reprodue asexually by BINARY FISSION mehanism. In this the genetic material is
replicated as same as the parent genome the cell devides in two parts and give rise to new
progeny so no geneti diversity is formed.
protists- they also reprodue asexually through binary fission, BUDDING In this new organism
grow from the body of parent.
PLANT- VEGETATIVE reproduction is the main asexual reproduction in which a plant part
give rise of new plant same as the parent, such as growing tips of RHIZOME.
ANIMALS- they also reproduce asexually through budding,fission,fragmentation.In case of stick
inset, bees PARTHENOGENESIS our in whih unfertilised egg give rise to new individual it is a
type of AGAMOGENESIS..
Define intermediate phenotype and then imagine some intermediate phe.pdf
Define intermediate phenotype and then imagine some intermediate phenotypes (e.g. RNA,
metabolite, etc) for Down’s Syndrome.
Solution
intermediate phenotype :
it depicts a living being that shows a mix of the phenotypic characteristics that its parents
expressed.
. In the event that a red blossom cross-pollinates a white bloom, and the subsequent offspring are
pink, the posterity portrayed as a intermediate phenotype .
It happen when the traits under investigation are neither dominant or recessive.
Down disorder (DS or DNS), otherwise called trisomy 21, is a hereditary issue brought about by
the nearness of all or some portion of a third duplicate of chromosome 21. It is commonly
connected with physical development delays, trademark facial elements and mellow to direct
learned incapacity.
Down disorder is brought on by having three duplicates of the qualities on chromosome 21, as
opposed to the standard two.The guardians of the influenced individual are ordinarily
hereditarily ordinary
Those who have one tyke with Down disorder have around a 1% danger of having a moment
youngster with the disorder, if both guardians are found to have typical karyotypes.
The additional chromosome substance can emerge through a few distinctive ways.
The most widely recognized cause (around 92–95% of cases) is an entire additional duplicate of
chromosome 21, bringing about trisomy 21. In 1.0 to 2.5% of cases, a portion of the phones in
the body are typical and others have trisomy 21, known as mosaic Down syndrome.
The other basic systems that can offer ascent to Down disorder include: a Robertsonian
translocation, isochromosome, or ring chromosome.
These contain extra material from chromosome 21 and happen in around 2.5% of cases. An
isochromosome comes about when the two long arms of a chromosome isolate together as
opposed to the long and short arm isolating together amid egg or sperm improvement.
Trisomy 21 is known by the karyotype 47,XX,+21 for females and 47,XY,+21 for guys) is
brought about by a disappointment of the 21st chromosome to isolate amid egg or sperm
development.As an outcome, a sperm or egg cell is delivered with an additional duplicate of
chromosome 21; this cell therefore has 24 chromosomes.
At the point when joined with a typical cell from the other parent, the child has 47 chromosomes,
with three duplicates of chromosome 21.
Around 88% of instances of trisomy 21 result from nonseparation of the chromosomes in the
mother, 8% from nonseparation in the father, and 3% after the egg and sperm have combined.
Translocation : The additional chromosome 21 material may likewise happen due to a
Robertsonian translocation in 2–4% of cases. In this circumstance, the long arm of chromosome
21 is joined to another chromosome, regularly chromosome 14.
In a male influenced with Down disorder, it brings about a karyotype of 46XY,t(14q21q). This
might be another transformation or already display in one of the guardians.
The parent with such a translocation i.
Consider a population of lizards living on the coast of Africa. A sto.pdf
Consider a population of lizards living on the coast of Africa. A storm creates piles of debris
that the lizards use to raft to a faraway uninhabited island. Which evolutionary process is
happening? mutation selection balance the founder effect frequency-dependent selection
coalescence
Solution
Answer is. \"Founder effect\".
The lizards completely moved to new place where they never I habited before, and hence they
found new environment to adjust, which will create some environmental difference. This is
known as founder effect..
Can “discovery science” (for example, the discovery of a new species.pdf
Can “discovery science” (for example, the discovery of a new species of fossil, or a new species
of organism that is alive today) contribute to our understanding of evolution? Give examples.
Solution
Scientists seek natural causes for natural phenomena. This is the most basic description of what a
scientist aims to do every day. Their focus is limited to what they can physically observe and
measure directly, sometimes with the help of tools such as a microscope. They do not seek to
explain the supernatural as it is outside the bounds of science, but instead what is all around
them. There are basic methods of gaining knowledge that are common to all of science.
Biology uses discovery science , and hypothesis-based science . While discovery science is
mostly about describing nature, hypothesis-based science attempts to explain nature. The pair of
inquiries are widely used together in research. Observations that can be measured and verified
are the data of discovery science. In the study of life, discovery science is used to describe both
the simplest of life, and the most complex. From the building blocks of atoms and molecules all
the way up to the ecosystems and biosphere, discovery science describes it all. Discovery
science is based on the logic of inductive reasoning . Inductive reasoning uses numerous specific
observations to make a general statement.
For example: all living things are made of cells. This comes after the numerous discoveries of
cells in every specimen observed by biologists over two centuries. another example we can take
\"humans\". today we found most of the fossils of human lives who lived thousands of years
back and with help of that fossils we can understand growth and evolution human life..
All of the following are features or functions of nanobodies except _.pdf
All of the following are features or functions of nanobodies except ______________. small,
monomeric, lack disulfide bonds, and resistance to denaturation high affinity to the antigen
recognize protruding and recessed paratopes cannot be humanized consist of only the VHH
domain of heavy chain antibodies
Solution
d)cannot be humanized
because we have many humanized antibodies that are available .example-CyaA-Hly specific
humanized antibody.
A girl running at a constant speed of 1.4ms in a straight line thro.pdf
A girl running at a constant speed of 1.4m/s in a straight line throw a ball directly upward at 4.7
m/s. How far will she travel before the ball drops back into her hand? Ignore air fiction.
Solution
First find out how long the ball will be in the air,
t = seconds
Gravitational constant: -4.9 m/sec^2
Upward velocity: +4.7 m/s
:
-4.9t^2 + 4.7t = 0
t(-4.9t + 4.7) = 0.
You have a rural dial-up customer who complains that a large number .pdf
You have a rural dial-up customer who complains that a large number of phone calls have been
made to “dangerous” sites on the Internet from his computer but he didn’t make these calls. You
investigate his computer and discover what type of malware?
A. Adware
B. Dialer
C. Keystroke logger
D. Spim
Solution
B. Dialers are types of malicious programs installed on computers that cause the dialup modem
to call dangerous or pay-per-call locations on the Internet, often resulting in extremely high
phone bills for the customer..
You are given a mixed culture containing a anaerobic thermophile, ae.pdf
You are given a mixed culture containing a anaerobic thermophile, aerobic thermophile and
anaerobic hyperthermophile bacteria. How will you separate these into pure cultures?
Solution
In natural environments, microorganisms usually exist as mixed populations. A pure culture is
one by which all organisms are descendants of the exact same organism. A sterile medium is the
one which is free of all life forms. It is often sterilized by heating it to a temperature at which all
contaminating microorganisms are destroyed. Eventually, in functioning with microorganisms,
we ought to have a way of transferring growing organisms from a pure culture to a sterile
medium without introducing any undesirable outside contaminants. This approach of stopping
unwanted microorganisms from developing accessibility is termed aseptic technique.
Microorganisms whose optimal temperature for development is near 0oC are called
psychrophiles. The ones that grow optimally between 20oC and 45eC are called mesophiles.
These growing most useful at large conditions, 55oC to ~ 70oC and larger, are called
thermophiles. Prokaryotes that live at quite high conditions (up to 113oC ) are
hyperthermophiles. An organism that will require air for development is called aerobic. An
organism whose development can\'t arise in the presence of air is anaerobic. An organism that
may grow under either aerobic or anaerobic conditions is a facultative anaerobe. Microaerophilic
organisms involve free air but at a limited concentration. Based on this, the plates are incubated
at different temperatures with aerobic and anerobic conditions for enhancing the growth of
thermophiles and hyperthermophilc bacteria..
Woyld removing phenylethyl alcohol from PEA alter the mediumsS.pdf
Woyld removing phenylethyl alcohol from PEA alter the medium\'s
Solution
Phenylethyl alcohol agar (PEA) medium is a selective media that inhibit the growth of gram-
negative bacteria and allows the growth of gram-positive bacteria only. The phenyl ethyl alcohol
is responsible for preventing the growth of gram-negative bacteria. So, removing phenyl ethyl
alcohol from PEA alter the medium\'s specificity againstgram positivebacteria..
Which abstraction uses slates, state transitions, inputs and outputs .pdf
Which abstraction uses slates, state transitions, inputs and outputs to model a complex system?
How is the order of the entries in a linked list determined?
Solution
7)
State Machines are used states, state transistion diagrams, and inpuit & output systems. State
Machines outputs are depending on the entire history if theire inputs..
What is the target cell for the hormone AngiotensionogenSolut.pdf
What is the target cell for the hormone Angiotensionogen?
Solution
Answer: zona glomerulosa cells of the adrenal cortex in the adrenal gland on kidney
Reason:
Primarily low blood pressure trigger the release of precursor, angiotensinogen peptide, a serpin
family from the liver cells followed by formation of angiotensin I finally to angiotensin II in the
presence of angiotensin converting enzyme (ACE). This angiotensin II active form is going to
act on the \"zona glomerulosa cells of the adrenal cortex in the adrenal gland on kidney\" to
trigger release of aldosterone to act on kidney cells finally to the reabsorption of sodium.
What is a possible evolutionary advantage of having intronsexons in.pdf
What is a possible evolutionary advantage of having introns/exons in genes? Give an example of
where this usage is observed or may have occurred.
Solution
Presence of large noncoding sequences such as introns interspersed with exons in the genome
lowers the rate mutations from appearing. Introns enable alternative splicing, a mechanism which
enables a single gene to encode multiple proteins that perform different functions under different
conditions. Small genome but large information stored. Regulatory RNAs such as siRNA
encoded by introns are known to regulate gene expression. Adds to biological complexity..
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Hilary rode her horse for 8 miles until it was hurt.Then she walked .pdf
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fully comments for my program, thank you will thumb up#include io.pdf
fully comments for my program, thank you will thumb up#include io.pdf
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C++Write a function void headEnqueue(Queue q, int key) which enqu.pdf
- 1. C++ Write a function void headEnqueue(Queue *q, int key) which enqueues the key such that, at function exit, it is positioned at the head of the queue rather than the tail; and the order and integrity of the queue is otherwise maintained. Solution struct node { int data; node* next; }; class Queue //linked_list class { private: node* front; node* rear; public: Queue(); void insert(int); headEnqueue(Queue *q, int key); }; Queue::Queue() { front = rear = NULL; } void Queue::insert(int val) { node* temp = new node; temp->data = val; temp->next = NULL; if (front == NULL) front= rear = temp;
- 2. else { node* temp1 = front; while (temp1->next) temp1 = temp1->next; temp1->next = temp; } } void Queue::headEnqueue(Queue *q, int key) { node* temp = new node; temp->data = key; temp->next=q->front; q->front=temp; }