1. With what data structure can a priority queue be implemented?
a) Array
b) List
c) Heap
d) Tree
Answer: c
Explanation: A priority queue can indeed be implemented using various data structures such as an array, a linked list, a binary search tree, or a heap. However, the most efficient implementation is typically the heap, specifically a binary heap, because it allows for faster insertion and removal of the highest-priority element, with both operations typically having a time complexity of O(log n).
2. Which of the following is not an application of priority queue?
a) Huffman codes
b) Interrupt handling in operating system
c) Undo operation in text editors
d) Bayesian spam filter
Answer: c
Explanation: Undo operations in applications are typically implemented using a stack data structure. This is because stacks follow the Last In, First Out (LIFO) principle, which allows you to easily revert to the most recent state by popping the top item of the stack. Each action (such as typing a letter or editing a document) is pushed onto the stack, and when an undo action is triggered, the last action is popped and reversed.
3. Select the appropriate code that inserts elements into the list based on the given key value.
(head and trail are dummy nodes to mark the end and beginning of the list, they do not contain any priority or element)
a)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
head.setNext(temp);
temp.setNext(trail);
}
else
{
Node dup = head.getNext();
Node cur = head;
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(temp);
temp.setNext(dup);
}
count++;
}
}
b)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
head.setNext(temp);
temp.setNext(trail);
}
else
{
Node dup = head.getNext();
Node cur = dup;
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(temp);
temp.setNext(dup);
}
count++;
}
}
c)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
head.setNext(temp);
temp.setNext(trail);
}
else
{
Node dup = head.getNext();
Node cur = head;
while((key>dup.getKey()) && (dup!=trail))
{
dup = dup.getNext();
cur = cur.getNext();
}
cur.setNext(dup);
temp.setNext(cur);
}
count++;
}
}
d)
public void insert_key(int key,Object item)
{
if(key<0)
{
Systerm.our.println("invalid");
System.exit(0);
}
else
{
Node temp = new Node(key,item,null);
if(count == 0)
{
head.setNext(temp);
temp.setNext(trail);
}
else
{
Node dup = head.getNext();
Node cur = head;
while((key>dup.getKey()) && (dup!=trail))
{
dup = cur
cur = cur.getNext();
}
cur.setNext(dup);
temp.setNext(cur);
}
count++;
}
}
Answer: a
Explanation: To insert a new node into a sorted linked list, we use two temporary pointers: dup and cur. The dup pointer is used to traverse the list, while cur trails behind dup. As we traverse the list, we compare the key with each element’s value. When we find an element greater than the key, we insert the new node temp in that position. This ensures the list remains sorted after the insertion.
4. What is the time complexity to insert a node based on key in a priority queue?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n2)
Answer: c
Explanation: In the worst-case scenario, to insert an element into a sorted linked list, you may need to traverse through the entire list to find the correct position for the new node. Therefore, the time complexity for this operation is O(n), where n is the number of elements in the list.
5. What is the functionality of the following piece of code?
public Object delete_key()
{
if(count == 0)
{
System.out.println("Q is empty");
System.exit(0);
}
else
{
Node cur = head.getNext();
Node dup = cur.getNext();
Object e = cur.getEle();
head.setNext(dup);
count--;
return e;
}
}
a) Delete the second element in the list
b) Return but not delete the second element in the list
c) Delete the first element in the list
d) Return but not delete the first element in the list
Answer: c
Explanation: In this case, a pointer is used to point to the first element in the list, and another pointer is used to point to the second element. By manipulating the pointers, the first element is effectively removed from the list, and its value is returned. This operation is typically performed when deleting the first element from a linked list, updating the head pointer to point to the second element.
6. What is not a disadvantage of priority scheduling in operating systems?
a) A low priority process might have to wait indefinitely for the CPU
b) If the system crashes, the low priority systems may be lost permanently
c) Interrupt handling
d) Indefinite blocking
Answer: c
Explanation: In a priority scheduling system, lower priority processes are held off until higher priority processes have finished executing. This ensures that more important tasks are completed first. Interrupt handling benefits from this mechanism because interrupts, which typically represent urgent tasks, are given higher priority than regular tasks. This allows the system to quickly respond to interrupts and handle them promptly, ensuring efficient processing and desired results.
7. Which of the following is not an advantage of a priority queue?
a) Easy to implement
b) Processes with different priority can be efficiently handled
c) Applications with differing requirements
d) Easy to delete elements in any case
Answer: d
Explanation: In a priority queue, the element with the highest priority is typically located at the front. However, in some implementations (such as a linear search or an unsorted list), finding the highest priority element for deletion may require scanning the entire queue. This results in a time complexity of O(n), which is less efficient compared to using a more advanced structure like a heap, where deletion of the highest priority element can be done in O(log n) time. Therefore, deleting elements in the worst case can take longer.
8. What is the time complexity to insert a node based on position in a priority queue?
a) O(nlogn)
b) O(logn)
c) O(n)
d) O(n2)
Answer: c
Explanation: In a priority queue, if it’s implemented using an unsorted list or array, finding the element with the highest priority might require traversing the entire list. In the worst case, this traversal takes O(n) time, where n is the number of elements in the queue. This is because each element must be compared to find the highest priority, making the operation less efficient than using a sorted structure like a heap.