Linked List Practice Questions

The C code iterates through L1 using ptr1. In each step, it takes the data of the next node in L1 (i.e., ptr1->next->data) and searches for this value in L2 using the find function.

If the find function returns true (meaning the value exists in L2), ptr1->next is updated to ptr1->next->next, effectively removing the node that contained the matched data from L1. If find returns false, ptr1 simply moves to ptr1->next.

Let's trace the execution:

1. ptr1 points to 1. query = 7. find(7, L2) is false. ptr1 moves to 7. L1: 1 -> 7 -> 12 -> 3 -> 9 -> 5 -> 11 -> 15 -> 8
2. ptr1 points to 7. query = 12. find(12, L2) is true. Node 12 is removed. L1: 1 -> 7 -> 3 -> 9 -> 5 -> 11 -> 15 -> 8
3. ptr1 points to 7. query = 3. find(3, L2) is false. ptr1 moves to 3. L1: 1 -> 7 -> 3 -> 9 -> 5 -> 11 -> 15 -> 8
4. ptr1 points to 3. query = 9. find(9, L2) is true. Node 9 is removed. L1: 1 -> 7 -> 3 -> 5 -> 11 -> 15 -> 8
5. ptr1 points to 3. query = 5. find(5, L2) is false. ptr1 moves to 5. L1: 1 -> 7 -> 3 -> 5 -> 11 -> 15 -> 8
6. ptr1 points to 5. query = 11. find(11, L2) is true. Node 11 is removed. L1: 1 -> 7 -> 3 -> 5 -> 15 -> 8
7. ptr1 points to 5. query = 15. find(15, L2) is true. Node 15 is removed. L1: 1 -> 7 -> 3 -> 5 -> 8
8. ptr1 points to 5. query = 8. find(8, L2) is false. ptr1 moves to 8. L1: 1 -> 7 -> 3 -> 5 -> 8
9. ptr1 points to 8. ptr1->next is NULL. Loop terminates.

The final list L1 is 1 -> 7 -> 3 -> 5 -> 8. There are 5 nodes remaining.

[CORRECT_OPTION: 5]

We want to find an integer in an unordered list that is NOT the largest.
Consider an arbitrary list of $N$ distinct integers.
To find an integer that is NOT the largest, we only need to compare two distinct elements.
Let the list be $L = \{e_1, e_2, \dots, e_N\}$.
Pick any two elements from the list, say $e_1$ and $e_2$.
Compare $e_1$ and $e_2$.
Case 1: $e_1 < e_2$. In this case, $e_1$ is guaranteed not to be the largest element in the list. So, $e_1$ is an answer.
Case 2: $e_2 < e_1$. In this case, $e_2$ is guaranteed not to be the largest element in the list. So, $e_2$ is an answer.
In either case, after one comparison, we can identify an element that is not the largest.
Therefore, the minimum number of element comparisons required is 1.

SLLdel (Singly Linked List deletion): To delete a node in a singly linked list, given a pointer to the node to be deleted, you need to find its predecessor to update the 'next' pointer. In the worst case, this requires traversing the list from the head, which takes $O(n)$ time.
DLLdel (Doubly Linked List deletion): To delete a node in a doubly linked list, given a pointer to the node to be deleted, you have access to both its predecessor and successor through 'prev' and 'next' pointers. You can update these pointers in constant time, $O(1)$.
Therefore, SLLdel is $O(n)$ and DLLdel is $O(1)$.

The problem is to reverse a singly linked list in $O(1)$ space.
(A) "The best algorithm for the problem takes $\theta(n)$ time in the worst case."
To reverse a singly linked list, you need to traverse it at least once to change pointers. Traversing an $n$-node list takes $\theta(n)$ time. Therefore, an algorithm for reversing the list must be at least $\theta(n)$ in time complexity.
(D) "It is not possible to reverse a singly linked list in $O(1)$ space."
This statement is FALSE. A standard iterative algorithm for reversing a singly linked list uses three pointers (previous, current, next), which is $O(1)$ additional space. The algorithm traverses the list, changing the next pointer of the current node to point to previous, then advancing the previous and current pointers.
Since (D) is false, and (A) describes the time complexity of the $O(1)$ space algorithm, (A) is true.
The iterative approach for reversing a singly linked list in $O(1)$ space takes $\theta(n)$ time because it iterates through all $n$ nodes once.

The program creates a linked list.
Initially, head and boxE point to a newly allocated node (value 0).
The first loop runs for index from 1 to 3 (3 iterations). In each iteration, a new node is created, its value is set to index, and it's appended to the list.
So, after the first loop, the list is: 0 -> 1 -> 2 -> 3. This is a linked list of 4 nodes.
The second loop runs for index from 0 to 3 (4 iterations).
printf("Value at index %d is %d\n", index, head->value); prints the value of the current head node.
head = head->next; moves head to the next node.

Iteration 1 (index=0): Prints "Value at index 0 is 0". head moves to node 1.
Iteration 2 (index=1): Prints "Value at index 1 is 1". head moves to node 2.
Iteration 3 (index=2): Prints "Value at index 2 is 2". head moves to node 3.
Iteration 4 (index=3): Prints "Value at index 3 is 3". head moves to NULL (since node 3's next was not explicitly set, it would be NULL or garbage).

The second printf statement printf("Value at index %d is %d\n", index+1, head->value); is problematic.
When index=3, head becomes NULL. The next printf attempts to dereference head->value when head is NULL. This will cause a run-time error (segmentation fault).
Therefore, the program dereferences an uninitialized pointer that may result in a run-time error.

To insert $n$ elements into an empty linked list and maintain sorted order, each insertion will involve traversing a portion of the existing list.

1. For the first element, insertion takes $\Theta(1)$ time.
2. For the second element, in the worst case, we might need to traverse the entire list (1 element) to find its sorted position, taking $\Theta(1)$ time.
3. For the $k$-th element, in the worst case, we might need to traverse the entire sorted list of $k-1$ elements to find its correct position, taking $\Theta(k-1)$ time.
4. The total worst-case time complexity for inserting $n$ elements is the sum of times for each insertion: $\sum_{k=1}^{n} \Theta(k-1) = \Theta(0 + 1 + \dots + n-1) = \Theta\left(\frac{(n-1)n}{2}\right) = \Theta(n^2)$.

For a singly linked list with a head pointer $F$ and a tail pointer $L$, deleting the last node requires updating the next pointer of the node preceding it to $NULL$ and updating $L$. Because the list is singly linked, there is no backward pointer to access the second-to-last node directly from $L$. Therefore, one must traverse the list starting from $F$ to find the node before $L$, which takes $O(n)$ time, where $n$ is the length of the list. In contrast, deleting the first node (B), adding an element after $L$ (C), or swapping the first two nodes (D) involves only constant time $O(1)$ operations as the necessary pointers are directly accessible or involve only the head of the list.

To delete a node $X$ from a doubly linked list, we must update the pointers of its adjacent nodes to bypass $X$ and link them directly to each other.

1. Update the predecessor: The node before $X$ is accessed via $X \rightarrow Bwd$. Its forward pointer, $X \rightarrow Bwd \rightarrow Fwd$, must point to the node after $X$, which is $X \rightarrow Fwd$. Thus: $X \rightarrow Bwd \rightarrow Fwd = X \rightarrow Fwd;$.
2. Update the successor: The node after $X$ is accessed via $X \rightarrow Fwd$. Its backward pointer, $X \rightarrow Fwd \rightarrow Bwd$, must point to the node before $X$, which is $X \rightarrow Bwd$. Thus: $X \rightarrow Fwd \rightarrow Bwd = X \rightarrow Bwd;$.

Option A correctly performs these two pointer assignments, effectively removing $X$ from the list structure.

A circular linked list allows for insertion and deletion at any arbitrary position, such as the beginning, the end, or after any specific node. Statement 1 is false because insertion is not restricted solely to the last node. Statement 2 is also false because deletion can occur at any position, not exclusively at the last node. Since both statements incorrectly define the standard operations of a circular linked list by imposing unnecessary restrictions, both are false.

In a doubly linked list, each node contains both a $next$ and $prev$ pointer. An insertion operation requires updating these pointers to maintain the integrity of the list, but the specific count depends on the location of the insertion. For example, inserting a node at the beginning involves updating the $head$ pointer and the $prev$ pointer of the first node. In contrast, inserting a node between two existing nodes ($A$ and $B$) typically requires updating four pointers: the new node's $next$ and $prev$, node $A$'s $next$, and node $B$'s $prev$. Since the number of affected pointers varies based on the insertion position and list state, it is not a fixed constant.

The join(m, n) function is designed to append the linked list m to the end of the linked list n.

The code first initializes a pointer p to the head of list n. The while loop then traverses list n until p points to the last node (i.e., the node whose next pointer is NULL). Finally, it sets the next pointer of this last node to m, linking the two lists.

However, there is a critical edge case. If the list n is empty, the input n will be NULL. In this scenario, p is initialized to NULL, and the subsequent attempt to access p->next will cause a null pointer dereference. Thus, the function either appends the list or causes this error.

The time complexity for each operation is analyzed as follows:

1. $\Theta(N)$ delete operations: Deleting a node from a doubly linked list given a pointer takes $O(1)$ time. Therefore, $\Theta(N)$ delete operations will take $\Theta(N) \cdot O(1) = \Theta(N)$ time.
2. $O(logN)$ insert operations: Inserting into a sorted doubly linked list requires finding the correct position, which takes $O(N)$ in the worst case. So, $O(logN)$ inserts take $O(logN) \cdot O(N) = O(NlogN)$ time.
3. $O(logN)$ find operations: Finding an element in a sorted doubly linked list takes $O(N)$ time in the worst case. Thus, $O(logN)$ find operations will take $O(logN) \cdot O(N) = O(NlogN)$ time.
4. $\Theta(N)$ decrease-key operations: A decrease-key operation on a sorted doubly linked list involves updating the key and then repositioning the node. Repositioning requires deleting the node ($O(1)$ with pointer) and then inserting it back ($O(N)$ to find the new position). Therefore, one decrease-key operation takes $O(N)$. For $\Theta(N)$ decrease-key operations, the total time will be $\Theta(N) \cdot O(N) = \Theta(N^2)$.

Combining all complexities: $\Theta(N) + O(NlogN) + O(NlogN) + \Theta(N^2)$. The dominant term is $\Theta(N^2)$.

The overall time complexity is $O(N^2)$.

The problem requires selecting the most suitable data structure based on the frequency of different operations to achieve the best total asymptotic complexity. The algorithm performs $N$ insertions, which is asymptotically a much larger number of operations compared to the $(\log N)^{1/2}$ find, delete, and decrease-key operations.

Because the number of insert operations dominates, the complexity of a single insertion is the most critical factor for the total time complexity. We compare the insertion costs:

• Unsorted array: $O(1)$
• Min-heap: $O(\log N)$
• Sorted array/list: $O(N)$

The unsorted array offers the lowest cost for the most frequent operation, making it the most suitable choice to optimize the overall performance.

To find an element that is neither the maximum nor the minimum, we only need to consider a small, constant number of elements from the list. Select any three distinct elements. By performing at most three comparisons on these elements, we can identify the one with the middle value. This middle element is guaranteed to be neither the overall maximum nor the overall minimum of the entire list. Since the number of operations is constant and does not depend on the list size $n$, the time complexity is $\Theta(1)$.

In a singly linked list, navigation is unidirectional from $F$. To delete the last node (at $L$), we must set the next pointer of the node preceding it to NULL. Since we cannot traverse backward from $L$, we must traverse the list starting from $F$ to locate this penultimate node. This search process requires $O(n)$ time, where $n$ is the number of elements. Conversely, operations like deleting the first element, swapping the first two elements, or adding an element at the end (using $L$) can be performed in constant $O(1)$ time. Thus, deleting the last element is the only operation whose time complexity depends on the length of the list.

The sequence $p = \text{getnode}()$ allocates a new memory block for a node. Assigning $\text{info}(p) = 10$ initializes the data value, while $\text{next}(p) = \text{list}$ links the new node to the current front of the list. Finally, updating $\text{list} = p$ makes the new node the head of the list. Since these steps create and link a new node to the structure, this is an insertion operation.

The goal of the function is to move the last node of a singly-linked list to the front. The provided code already handles edge cases (empty or single-node list) and includes a while loop to traverse the list.

When the while loop finishes:

• p points to the last node of the list.
• q points to the second-to-last node (i.e., the node before p).
• head still points to the original first node.

To move p to the front, three steps are required:

1. Make the second-to-last node the new tail: The next pointer of q must be set to NULL. This detaches p from the list and makes q the new end. This is achieved by q->next = NULL;.

2. Make the old last node the new head: The next pointer of p should now point to the original head of the list. This is done with p->next = head;.

3. Update the list's head pointer: The head of the list must be updated to point to p, which is now the new first node. This is done with head = p;.

The correct sequence of these three statements is essential. The statement q->next = NULL; must be executed to make q the new tail. Then, p->next = head; links p to the front, and head = p; updates the head pointer.

Therefore, the correct code to fill the blank is q->next = NULL; p->next = head; head = p;.

In a singly linked list, elements are stored in nodes that are linked sequentially. To search for a specific value, one must perform a linear traversal starting from the head node. In the worst-case scenario, the target element is located at the last position or is not present, requiring the examination of all $n$ nodes. Since each node inspection is an $O(1)$ operation, the total time complexity is proportional to $n \times O(1)$, which results in $O(n)$.

The function iterates through the linked list, swapping the values of adjacent pairs of nodes. With the initial list $1 \to 2 \to 3 \to 4 \to 5 \to 6 \to 7$:

1. First pair (node 1, node 2): Swapping values results in $2 \to 1 \to 3 \to 4 \to 5 \to 6 \to 7$. Pointers $p$ and $q$ advance to nodes 3 and 4.
2. Second pair (node 3, node 4): Swapping values results in $2 \to 1 \to 4 \to 3 \to 5 \to 6 \to 7$. Pointers advance to nodes 5 and 6.
3. Third pair (node 5, node 6): Swapping values results in $2 \to 1 \to 4 \to 3 \to 6 \to 5 \to 7$.
4. The pointer $p$ moves to node 7, and $q$ becomes $NULL$ as there is no next node, terminating the loop. The final order is $2, 1, 4, 3, 6, 5, 7$.