 # Multiple Choice Questions on Linked Lists

### Understanding Linked Lists in Data Structures

A Linked List is a type of linear data structure similar to arrays. In a Linked List, each element is a separate object linked to the next using a reference stored in a pointer.

A Linked List is made up of nodes that are connected to the next node using memory references or addresses. Each Linked List node contains two properties:

• The value to be stored in the current node.
• The reference or address of the next node, or NULL if it's the last node.

There are two common types of Linked Lists:

• Singly Linked Lists: This is the standard Linked List where only the value of the current node and the reference of the next node is stored. The last node of the list has NULL stored as its reference.
• Doubly Linked Lists: This Linked List type stores all the information of the Singly Linked Lists, but also includes the reference or memory address of the previous node if it exists. Otherwise, the reference is NULL.

Below are some multiple-choice questions to test your understanding of Linked Lists:

### Information Stored in Nodes of a Doubly-Linked List

In a doubly-linked list, each node stores information about the value it holds, as well as the addresses of both the previous and the next nodes. Therefore, option D, which mentions all of the above, is correct.

### Optimal Time Complexity to Count Nodes in a Linked List

In a linked list, to count the number of nodes, we need to traverse through each node and increment a counter variable until the last node is reached. Thus, the time complexity is directly proportional to the number of nodes in the linked list.

Therefore, the optimal time complexity to count the number of nodes in a linked list is O(n), where n is the number of nodes in the linked list. Option A is the correct answer.

### Output of solve() function for the given linked list

The output of the given code snippet for the linked list 1->2->3->4->5->6 will be:

1 3 5 5 3 1

Explanation:

The given function solve() takes a pointer to the first node of the linked list as input and recursively prints alternative nodes of the list from start to end and then from end to start.

For the given linked list 1->2->3->4->5->6, the output of the solve() function will be:

- start->data is 1, so print 1 - The next nodes are 2 and 3, so recursively call solve() with the second next node i.e. start->next->next i.e. node 3. - node 3's data is 3, so print 3 - The next nodes are 4 and 5, so recursively call solve() with the second next node i.e. start->next->next i.e. node 5. - node 5's data is 5, so print 5. - The next node is NULL, so return. - At this point, the first half of the list has been printed in order: 1 3 5. - Now, we are in the second half of the list and start->data is still 5, so print 5. - The next node is 4, but its data is already printed in the first half. So, jump to the next node i.e. 3. - 3's data is printed earlier, so jump to 2. - 2's data was not printed in the first half, so print 2. - The next node is NULL, so return. - Now, the second half of the list has been printed in reverse order: 5 3 1. - Hence, the final output is: 1 3 5 5 3 1.

Linked Lists have various applications in computer science, including:

1. Implementing file systems
2. Chaining in hash tables
3. Binary Trees implementation

Therefore, the correct answer is D) All of the above, as Linked Lists are used in all of these processes.

### Inserting an Element in the Middle of a Linked List

When inserting an element in the middle of a linked list, two pointers need to be modified - the pointer pointing to the node before the node to be inserted, and the pointer pointing to the node being inserted.

``// Example code for inserting a node in the middle of a linked list``
``public void insertInMiddle(Node prevNode, Node newNode) {``
``    newNode.next = prevNode.next;``
``    prevNode.next = newNode;``
``}``

### Inserting an Element at the Ends of a Linked List

When an element needs to be inserted at the ends of a linked list, only 1 pointer needs to be modified.

### Worst Case Comparison in Singly Linked List Search

When searching for an element in a singly linked list of length n, the worst case scenario occurs when the element being searched for is at the end of the list or not in the list at all. In this case, we would need to traverse the entire list to determine that the element is not present. Therefore, the number of comparisons needed in the worst case is O(n).

### Feasibility of Implementing Algorithms in a Linked List

In a linked list, not all algorithms can be implemented feasibly due to the way data is stored. The following algorithms can be efficiently implemented in a linked list:

• Linear Search
• Merge Sort
• Insertion Sort

However, the Binary Search algorithm is not feasible to implement in a linked list due to the inability to directly access an element at a random location.

``Answer: D)``

### Calculate Sum in a Linked List

Given a linked list of integers, we want to calculate the sum of all the integer values found in the linked list. Here's a code snippet that implements a helper function for calculating the sum:

``````
/*
head-> val = value of the node
The List is 1 -> 2 -> 3 -> 4 -> 5
*/
int sum = 0;
}
cout << "The sum of the given linked list is: " << sum << endl;
}
``````

To use this helper function, simply pass the head node of your linked list as a parameter:

``````
``````

After running the given code snippet, the value of "sum" will be the sum of all values of the linked list.

A LinkedList can be used for various operations including:

``1. Implementation of Stacks and Queues. ``
``2. Implementation of Binary Trees. ``
``3. Implementation of Data Structures that can simulate Dynamic Arrays.``

Therefore, the correct answer is: All of the above.

### Time complexity to insert an element to the front of a LinkedList

The time complexity to insert an element to the front of a LinkedList (given the head pointer) is O(1). This is because we only need to change the reference of the head to the new node and make the new node point to the previous head. This operation takes a constant amount of time, regardless of the size of the LinkedList.

Here's some sample code:

``````code
Node newNode = new Node(data); // create the new node
newNode.next = head; // point the new node to the current head
``````

Note that this assumes the LinkedList implementation has constant-time access to the head pointer. In some cases, such as with a doubly linked list, inserting an element at the front may require traversal of the list, resulting in a different time complexity.

### Time Complexity to Insert an Element to the Rear of a Linked List

The time complexity to insert an element to the rear of a linked list when head pointer is given is O(n).

The reason for this is that we have to traverse through the linked list until we reach the end in order to insert the new element at the rear. This traversal takes linear time and is directly proportional to the size of the linked list, hence resulting in a time complexity of O(n).

Therefore, option A is the correct answer.

### Implementing a Linked List Node in C++

In C++, linked list nodes can be implemented using both structs and classes.

``````
// Defining a struct for a linked list node
struct Node {
int data;
Node* next;
};

// Defining a class for a linked list node
class Node {
public:
int data;
Node* next;
};
``````

Both of these implementations allow creating a linked list by connecting multiple nodes.

In a Circular Linked List, the last node points to the head node, thereby forming a loop. This type of linked list stores the pointer of the head node in the next pointer of the last node.

A Singly Linked List only has one pointer which points to the next node. A Doubly Linked List on the other hand has two pointers which point to the previous and next nodes. A Hashed List is a type of linked list that uses a hash function to map values to their corresponding indices in the list.

``//Sample code for a Circular Linked List implementation``
``````python
#Creating a Node class
class Node:
def __init__(self, data):
self.data = data
self.next = None

#Creating a Circular Linked List Class
def __init__(self):
self.last = None

#Adding a node to the end of the list
#Create a new node
new_node = Node(data)
#Make it the last node
self.last = new_node
#Point it to itself
self.last.next = self.last

#Adding a node to the end of the list
#If list is empty
if self.last == None:
return
#Create a new node
new_node = Node(data)
#Point the last node to the new node
new_node.next = self.last.next
self.last.next = new_node
#Make the new node the last node
self.last = new_node

#Traversing the list
def traverse(self):
#If the list is empty
if self.last == None:
print("List is empty")
return
#Print the first node
print(self.last.next.data)
n = self.last.next.next
#Print the rest of the nodes
while n != self.last.next:
print(n.data)
n = n.next
``````

Code:

``````
#include <stdio.h>
#include <stdlib.h>

struct Node {
int coefficient;
int exponent;
struct Node *next;
};

// Function to create a new node
struct Node *createNode(int coef, int exp)
{
struct Node *temp = (struct Node *)malloc(sizeof(struct Node));
temp->coefficient = coef;
temp->exponent = exp;
temp->next = NULL;
return temp;
}

// Function to add two polynomials
struct Node *addPolynomials(struct Node *poly1, struct Node *poly2)
{
struct Node *result = NULL, *temp, *prev;
while (poly1 != NULL && poly2 != NULL) {
if (poly1->exponent > poly2->exponent) {
temp = createNode(poly1->coefficient, poly1->exponent);
poly1 = poly1->next;
}
else if (poly1->exponent < poly2->exponent) {
temp = createNode(poly2->coefficient, poly2->exponent);
poly2 = poly2->next;
}
else {
temp = createNode(poly1->coefficient + poly2->coefficient, poly1->exponent);
poly1 = poly1->next;
poly2 = poly2->next;
}
if (result == NULL)
result = prev = temp;
else {
prev->next = temp;
prev = temp;
}
}
while (poly1 != NULL) {
temp = createNode(poly1->coefficient, poly1->exponent);
poly1 = poly1->next;
prev->next = temp;
prev = temp;
}
while (poly2 != NULL) {
temp = createNode(poly2->coefficient, poly2->exponent);
poly2 = poly2->next;
prev->next = temp;
prev = temp;
}
return result;
}

// Function to print the polynomial
void printPolynomial(struct Node *poly)
{
while (poly != NULL) {
printf("%dx^%d ", poly->coefficient, poly->exponent);
poly = poly->next;
if (poly != NULL)
printf("+ ");
}
}

// Driver code
int main()
{
struct Node *poly1, *poly2, *result;

// Create first polynomial 3x^2 + 5x^1 + 6x^0
poly1 = createNode(3, 2);
poly1->next = createNode(5, 1);
poly1->next->next = createNode(6, 0);

// Create second polynomial 6x^1 + 8x^0
poly2 = createNode(6, 1);
poly2->next = createNode(8, 0);

printf("First polynomial: ");
printPolynomial(poly1);

printf("\nSecond polynomial: ");
printPolynomial(poly2);

printf("\nResultant polynomial: ");
printPolynomial(result);

return 0;
}
```
Explanation: The program implements polynomial addition using linked lists. It defines a Node struct with two integer values, coefficient and exponent, and a pointer to the next node in the list. The program uses the createNode function to create a new node with the given coefficient and exponent values, and returns that node. The addPolynomials function takes in two polynomial linked lists and adds them together to return a new linked list representing the sum. The printPolynomial function simply prints the polynomial in a user-friendly format. The main function creates two polynomial linked lists and prints them before and after adding them together using the addPolynomials function.
Explanation:
The given code snippet is used to reverse a singly linked list. It initializes a pointer "prev" to NULL. If the linked list is empty or has one node, it returns the linked list. Otherwise, it initializes a pointer "curr" to the second node and then iterates through the linked list in a loop. In each iteration, the "next" pointer of the current node is set to point to the previous node, then the "prev" pointer is updated to become the current node, the "head" pointer is set to become the "curr" pointer, and the "curr" pointer is updated to point to the next node. After the loop completes, the "prev" pointer points to the head of the reversed linked list, which is returned.Time Complexity of Reversing a Linked List
The time complexity of a program to reverse a linked list is O(n), where n is the number of nodes in the linked list. This is because in order to reverse the linked list, we need to traverse through each node and update the pointers accordingly.

Therefore, the correct answer is A) O(n).
Time Complexity of Concatenating Two Linked Lists
When concatenating two linked lists, the time complexity will be O(1) if we have the address of the last node of one of the lists. This is because we only need to update the next pointer of the last node of the first list to point to the head of the second list.
Therefore, the correct answer is option C) O(1) if we have the address of the last node of one of the lists.Types of Linked Lists that support bidirectional traversal
In Doubly Linked Lists, traversals can be performed in both forward and backward directions. This is because each node in a doubly linked list contains two pointers: one that points to the next node and another that points to the previous node. Singly Linked Lists, on the other hand, only have a pointer to the next node, so traversals can only be performed in the forward direction.
Optimal Algorithm to Find the Middle Element of a Linked List
The most efficient way to find the middle element of a linked list in terms of time complexity is by using the fast and slow pointer method. This algorithm involves using two pointers, a slow pointer and a fast pointer. The slow pointer moves one node per iteration, while the fast pointer moves two nodes per iteration. When the fast pointer reaches the end of the list, the slow pointer will be at the middle element.
Example code in Python:
```

while fast_ptr is not None and fast_ptr.next is not None:
fast_ptr = fast_ptr.next.next
slow_ptr = slow_ptr.next

return slow_ptr.data
```
In the above implementation, `head` is the first node of the linked list. The algorithm uses two pointers, `slow_ptr` and `fast_ptr`, both initially set to `head`. The loop continues until `fast_ptr` reaches the end of the list. The `slow_ptr` is the middle element of the list and is returned at the end of the function.Identifying a Linked List Without Null Pointers
In a linked list, each node contains two fields: a data field that stores the element and a pointer field that stores the reference to the next node in the list. A null pointer in the pointer field denotes the end of the list. However, if none of the nodes in the linked list points to null, it is a circular linked list.
A circular linked list connects all the nodes in a cyclic fashion without any null pointers. Hence option C - Circular Linked List is the answer to the given question.Checking the Truth of Statements about Linked Lists
The following statements are true regarding linked lists:
```
- Random access of elements at a linked list is not possible.
- Arrays have better cache locality than linked lists.
- The size of linked list is dynamic and can be changed as needed.
```
` D) All of the above statements are true.`

Optimal Complexity for Removing Duplicates from an Unsorted Linked List
The optimal complexity we can achieve when removing duplicates from an unsorted linked list is O(n). This can be done by using a hashing-based approach, which involves iterating through the linked list, adding each element to a hash table, and checking if an element already exists in the hash table before inserting it.
Therefore, the best approach to remove duplicates from an unsorted linked list is to use a hash table, which achieves linear time complexity.
Optimal Complexity for Removing Duplicates from a Sorted Linked List
The optimal complexity for removing duplicates from a sorted linked list is O(n). This means that the time it takes to remove duplicates from a linked list is proportional to the number of elements in the list.
This is achievable because the list is already sorted, which allows for a simple and efficient algorithm to remove duplicates. We can iterate through the list once and compare each element with the next element in the list. If they are the same, we can remove one of them.
This algorithm has a time complexity of O(n) because we only need to iterate through the list once. In contrast, if the list was not sorted, we would need to use a more complex algorithm that has a time complexity of O(n*logn) or O(n^2) to remove duplicates.

The following sorting algorithms can be applied to linked lists:
```
• Merge Sort
• Insertion Sort
• Quick Sort

However, Quick Sort is not a preferred choice for linked lists since linked lists do not support random access of elements.

### Code Snippet's Purpose

The given code snippet contains a function called `solve` that iterates through a Linked List pointed to by the head, incrementing the value of count by 1 each time, until it reaches NULL. The output of the value of count is then printed through cout.

There is a missing part of the code snippet, as the code ends abruptly after the `cout` statement. It is possible that the rest of the function's implementation is located elsewhere in the source code.

### Code Output Prediction for Given Input Linked List

Assuming the missing code is meant to print the value of the current node in each iteration of the while loop, the output for the input linked list 1->2->3->4->5 would be:

1 2 3 4 5

Code:

``````c++
cout << head->val << endl; // prints current node value
}
}
``````

### Type of Pointer Used to Point to the Address of the Next Element in a Linked List

In a linked list, the next node's address is stored in a pointer of node variable. Therefore, the correct answer is option C) Pointer to node. Pointers to character or integer cannot be used to point to the next element in a linked list.

### Code Snippet to Find Middle Element in a Linked List

This code snippet is used to find the middle element of a linked list.

``````
// Traverse the linked list with faster and slower pointers to find the middle element
while(fast->next != NULL && fast->next->next != NULL) {
fast = fast->next->next;
slow = slow->next;
}
// Return the middle element
return slow->val;
}
``````

The function takes the head pointer of the linked list as an argument and returns the value of the middle element. It uses two pointers, a fast pointer that moves two nodes at a time and a slow pointer that moves one node at a time. When the fast pointer reaches the end of the linked list, the slow pointer will be pointing to the middle element.

### Optimal way to find an element at the kth position in a Linked List

To find an element at the kth position in a Linked List, the optimal approach is to use the array implementation of the Linked List.

In this implementation, the elements of the Linked List are stored in an array. To find the element at the kth position, we just need to access the element at the k-1 index of the array, which is O(1) time complexity.

On the other hand, finding an element at the kth position in a Singly Linked List, Doubly Linked List, or Circular Linked List requires traversing through the list one by one until we reach the kth element. This approach takes O(n) time complexity, which is not optimal.

Therefore, the Array implementation of the Linked List is the optimal way to find an element at the kth position in a Linked List.The given code snippet modifies the pointer to the next node of the current node in the linked list. It does not delete any node from the linked list. Therefore, the correct answer is option A which states that the code deletes the current node from the linked list.

The time complexity of adding two linked list is O(max(n, m)), where n and m are the lengths of the two linked lists, respectively. This implies that the time it takes to add two linked lists is proportional to the maximum length of the two lists.

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode dummy(0);
ListNode* tail = &dummy;
int carry = 0;
while (l1 || l2 || carry) {
int val = carry;
if (l1) {
val += l1->val;
l1 = l1->next;
}
if (l2) {
val += l2->val;
l2 = l2->next;
}
tail->next = new ListNode(val % 10);
carry = val / 10;
tail = tail->next;
}
return dummy.next;
}
};``````

The above code is an example of adding two linked lists in C++. The solution implements an approach that adds two numbers digit by digit, starting from the least significant digit and propagating the carry forward. The function addTwoNumbers takes two linked lists (l1 and l2) as input and returns a linked list as output.

The time complexity of this implementation is O(max(n, m)), where n and m are the lengths of the two input lists, respectively. The space complexity is O(max(n, m)), since we create a new linked list that is at most the length of the larger of the two input lists.

### Solving problems with 2 Pointers on Linked Lists

Two pointers can be used on linked lists in an optimal way to solve the following problems:

1. Detecting cycle in a linked list
2. Finding intersection of 2 linked lists
3. Finding middle element of a linked list

Therefore, the correct answer is D) All of the above.

### Similarities between Singly and Doubly Linked Lists

In both singly and doubly linked lists:

• Data at a random position cannot be accessed in constant time
• A new node can be added after a given node or at the beginning of the list in O(1) time
• The first node can be deleted in O(1) time

Thus, option D is correct - all of the above statements are true for both singly and doubly linked lists.

``// Example code for adding a new node after a given node in singly linked list``
``````python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

def __init__(self):

if node is None:
print("Node does not exist.")
return
newNode = ListNode(value)
newNode.next = node.next
node.next = newNode
``````
``// Example code for deleting the first node in doubly linked list``
``````python
class ListNode:
def __init__(self, val=0, prev=None, next=None):
self.val = val
self.prev = prev
self.next = next

def __init__(self):

def deleteFirstNode(self):
print("List is empty.")
return
temp = None
``````

### Time Complexity of Rotating a Linked List by Some Places Clockwise

The time complexity of rotating a linked list by some places clockwise would be O(n).

The reason for this is that we need to traverse through the linked list up to the node where we want to rotate. This traversal would take O(n) time. Once we reach the node, we need to update the pointers of the nodes, which takes constant time. Therefore, the overall time complexity would be O(n).

### Space complexity required to rotate a linked list by some places clockwise

The space complexity required to rotate a linked list by some places clockwise is O(1).

``````
//function to rotate the linked list by k places clockwise
{
if (k == 0)
return;

int count = 1;
while (count < k && current != NULL) {
current = current->next;
count++;
}

if (current == NULL)
return;

Node* kthNode = current;

while (current->next != NULL)
current = current->next;

kthNode->next = NULL;
}
``````

The given implementation of the function "rotateListClockwise" only takes O(1) space complexity as it uses a constant amount of extra memory to change the links of the nodes in the linked list and doesn't create any new node.

### Time complexity of Enqueue operation in a Queue implemented by a Singly Linked List

The time complexity of Enqueue operation in a Queue implemented by a Singly Linked List is: O(1).

This means that the Enqueue operation will take constant time, regardless of the length of the queue.

Therefore, the correct answer is option A) O(1).

### Time Complexity of Dequeue Operation in a Queue Implemented by a Singly Linked List

In a singly linked list implementation of a queue, the dequeue operation has a time complexity of O(n).

``````
// Node class
class Node:
def __init__(self, data):
self.data = data
self.next = None

# Queue class
class Queue:
def __init__(self):
self.front = None
self.rear = None

def enqueue(self, data):
new_node = Node(data)

if self.rear is None:
self.front = self.rear = new_node
return

self.rear.next = new_node
self.rear = new_node

def dequeue(self):
if self.front is None:
return

temp_node = self.front
self.front = temp_node.next

if self.front is None:
self.rear = None
return temp_node.data
``````

The above code shows a Python implementation of a singly linked list-based queue. The dequeue operation, which is implemented in the dequeue() method, has a time complexity of O(n) since it requires iterating through the list to find the node just before the front node.

### Symbol Table: A Data Structure Used in Compilers

In a compiler, a symbol table is used to manage information about variables and their attributes. The symbol table stores all the variables' information such as name, data type, size, scope, and other relevant attributes. The symbol table makes it easy for the compiler to access pertinent information required for translation.

There are various ways to implement a symbol table, depending on the specific compiler design. However, the most common method is to use a hash table. The hash table allows for quick search and insertion of key-value pairs, where the key is the identifier and the value is the information about the identifier.

In conclusion, the use of a symbol table for managing information about variables and their attributes is an essential step in compiler design and helps to facilitate efficient and accurate translation of programming code.

In the context of polynomial addition, linked lists are typically used as a data structure. This is because a polynomial can be represented as a linked list where each node contains the coefficient and power of a term. Linked lists are a good fit for this purpose because they allow for efficient insertion and deletion of nodes. When adding two polynomials, the addition can be performed item-wise by traversing the two respective linked lists in parallel.

### Modifying Pointers in a Circular Linked List

In a circular linked list, inserting a record requires the modification of 2 pointers.

``````// code demonstrating insertion of a record in a circular linked list
void insertRecord(int data)
{
Node* newNode = new Node();   //create a new node
newNode->data = data;   //set the data of the new node to the given data
Node *temp = head;   //set a temp pointer to the current head

if (head == NULL)   //check if the list is empty
{
newNode->next = newNode;   //set the newNode to be the first and only node
}
else
{
while (temp->next != head)   //traverse to the last node
temp = temp->next;
temp->next = newNode;   //set the last node to point to newNode
newNode->next = head;   //set the newNode to point to head, thus making it circular
}
}``````

### Creating a new node in C programming language

In the given code snippet, a struct named "node" is defined with two members: "data" and "next". A typedef named "NODE" is created to define a type alias for the struct "node". A variable "ptr" of type "NODE*" is also declared.

To create a new node, the correct syntax is:

``````
ptr = (NODE*) malloc(sizeof(NODE));
``````

This allocates memory for a new node using the "malloc" function. The "sizeof" operator is used to calculate the size of the "NODE" struct, and the "ptr" pointer is set to the address of the allocated block of memory.

Option B is incorrect because it does not include the "sizeof" operator and would allocate memory for only the size of a pointer to a "NODE" struct, not the size of the struct itself.

Option C is incorrect because it adds an extra level of indirection by allocating memory for a pointer to a "NODE" struct, not the struct itself.

Option D is incorrect because it tries to cast a pointer to a struct, which is invalid.

### Destroying a Pointer in C++

In C++, you can destroy a pointer using the

``delete``

keyword. This keyword frees up the memory allocated for that pointer and releases any resources associated with it.

For example, suppose you have a pointer named

``ptr``

pointing to a dynamically allocated object. You can destroy it using the following code:

``delete ptr;``

You should always destroy pointers when you no longer need them to avoid memory leaks and other errors in your program.

Do not use

``free``

,

``calloc``

, or

``malloc``

to destroy C++ pointers, as they are used for dynamic memory allocation in C and can cause undefined behavior when used incorrectly in C++.

### True/False Quiz: Keywords and Library Functions

``````
// Check if the following statements are true or false

// Statement A: delete is a keyword.
// Statement B: free is a library function.

// Check if statement A is true
// by trying to declare a variable with the name "delete"
let delete = "This won't work";
// Result: SyntaxError - "delete" is a reserved word

// Check if statement B is true
// by using the "free" function to deallocate memory previously
// allocated with malloc or calloc
let arr = malloc(5 * sizeof(int));
free(arr);
// Result: The memory allocated to arr is deallocated and freed

// Therefore, statement A and statement B are both true
``````

### Space Complexity for Deleting a Linked List

In deleting a linked list, the space complexity is O(1). This means that we only need to maintain a temporary variable to keep track of the current node to be deleted, and no additional space is required.

### Space Complexity for Linked List

In order to store a linked list of n nodes, we need a space complexity of O(n). This means that the amount of space required to store the linked list will increase linearly as the number of nodes in the list increases.

The space complexity for a linked list is O(n) because each node in the list requires a certain amount of space to store its data and a pointer to the next node in the list. As the number of nodes increases, the amount of space required to store the linked list also increases proportionally.

Therefore, the correct answer is B) O(n).

### Linked List Operation with O(1) Time Complexity

The operation that takes O(1) time complexity in a linked list is:

``Inserting an element at the start of the linked list``

Insertion at the beginning of a linked list is faster as it doesn't require to traverse through the list to find the position to insert. The new element is simply added at the beginning, and the current head of the list becomes its next node.

In contrast, insertion at the end of a linked list takes O(n) time complexity as we have to traverse the list till the end to insert the new element.

### Information stored by linked lists for implementing binary trees

In a linked list used for implementing a binary tree, the following information is stored in each node:

- Value of the current node - Pointer to the left child - Pointer to the right child

Therefore, the correct answer is option D, which states that all the above information is stored.

### Optimizing Linked Lists for Searching in Better than O(N) Amortized Complexity

The most suitable variation of linked lists for searching in better than O(N) amortized complexity is the Skip List. It is specifically designed to allow for efficient search operations on a sorted list.

Other variations of linked lists such as singly linked lists or circular linked lists don't provide a significant advantage in terms of searching through a sorted list.

Therefore, if the goal is to optimize linked lists for efficient searches on a sorted list, then Skip Lists are the ideal choice.

### Preferred Sorting Algorithm for Linked List

When it comes to sorting a linked list, the preferred sorting algorithm is Merge Sort. Compared to other sorting algorithms, Merge Sort is preferred because it accesses the data sequentially and requires low random access. Quick Sort and Insertion Sort are less efficient than Merge Sort when it comes to sorting linked lists. Therefore, the answer is A) Merge Sort. ### Technical Interview Guides

Here are guides for technical interviews, categorized from introductory to advanced levels. ### Best MCQ

As part of their written examination, numerous tech companies necessitate candidates to complete multiple-choice questions (MCQs) assessing their technical aptitude.

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