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DSA C#programming~10 mins

Insert at Beginning of Doubly Linked List in DSA C# - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a new node with the given data.

DSA C#
Node newNode = new Node([1]);
Drag options to blanks, or click blank then click option'
Adata
Bvalue
CnodeData
Ditem
Attempts:
3 left
💡 Hint
Common Mistakes
Using a variable name that is not defined like 'value' or 'item'.
Forgetting to pass any argument to the Node constructor.
2fill in blank
medium

Complete the code to link the new node's next pointer to the current head.

DSA C#
newNode.[1] = head;
Drag options to blanks, or click blank then click option'
Aprev
Blink
Cnext
Dpointer
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'prev' instead of 'next' for the forward link.
Using an undefined property like 'link' or 'pointer'.
3fill in blank
hard

Fix the error in setting the previous pointer of the old head node.

DSA C#
if (head != null) head.[1] = newNode;
Drag options to blanks, or click blank then click option'
Apointer
Bprev
Cnext
Dlink
Attempts:
3 left
💡 Hint
Common Mistakes
Setting the old head's next pointer instead of prev.
Using an incorrect property name like 'pointer' or 'link'.
4fill in blank
hard

Fill both blanks to update the head pointer and the new node's previous pointer.

DSA C#
head = [1];
newNode.[2] = null;
Drag options to blanks, or click blank then click option'
AnewNode
Bhead
Cprev
Dnext
Attempts:
3 left
💡 Hint
Common Mistakes
Setting head to old head instead of new node.
Setting newNode.next to null instead of newNode.prev.
5fill in blank
hard

Fill all three blanks to complete the InsertAtBeginning method for a doubly linked list.

DSA C#
public void InsertAtBeginning(int data) {
    Node newNode = new Node([1]);
    newNode.[2] = head;
    if (head != null) head.[3] = newNode;
    newNode.prev = null;
    head = newNode;
}
Drag options to blanks, or click blank then click option'
Adata
Bnext
Cprev
Dpointer
Attempts:
3 left
💡 Hint
Common Mistakes
Mixing up next and prev pointers.
Not updating head to the new node.
Forgetting to set newNode.prev to null.

Practice

(1/5)
1. What is the main purpose of inserting a node at the beginning of a doubly linked list?
easy
A. To add a new node as the first element of the list
B. To remove the last node from the list
C. To search for a node in the list
D. To reverse the entire list

Solution

  1. Step 1: Understand the operation

    Inserting at the beginning means adding a new node before the current first node.
  2. Step 2: Identify the effect on the list

    The new node becomes the head (first node), shifting others to the right.
  3. Final Answer:

    To add a new node as the first element of the list -> Option A
  4. Quick Check:

    Insert at beginning = Add first node [OK]
Hint: Insert at beginning means new node becomes the first [OK]
Common Mistakes:
  • Confusing insertion with deletion
  • Thinking insertion adds at the end
  • Mixing up search with insertion
2. Which of the following is the correct syntax to create a new node in C# for a doubly linked list?
easy
A. Node newNode = new Node(data);
B. newNode = Node(data);
C. Node newNode = Node(data);
D. Node newNode = new Node[];

Solution

  1. Step 1: Recall C# object creation syntax

    In C#, objects are created using the 'new' keyword followed by the constructor call.
  2. Step 2: Match syntax with node creation

    Correct syntax is 'Node newNode = new Node(data);' to create a new node with data.
  3. Final Answer:

    Node newNode = new Node(data); -> Option A
  4. Quick Check:

    C# new object = new ClassName() [OK]
Hint: Use 'new' keyword to create objects in C# [OK]
Common Mistakes:
  • Omitting 'new' keyword
  • Using array syntax for single node
  • Missing type declaration
3. Consider this code snippet inserting a node with value 10 at the beginning of a doubly linked list:
Node newNode = new Node(10);
newNode.next = head;
if (head != null) {
    head.prev = newNode;
}
head = newNode;
What will be the printed list after insertion if the original list was: 20 <-> 30 <-> null?
medium
A. 30 <-> 20 <-> 10 <-> null
B. 20 <-> 30 <-> 10 <-> null
C. 10 <-> 30 <-> 20 <-> null
D. 10 <-> 20 <-> 30 <-> null

Solution

  1. Step 1: Insert new node at beginning

    New node with 10 points next to old head (20). Old head's prev points back to new node.
  2. Step 2: Update head pointer

    Head now points to new node (10), so list starts with 10.
  3. Final Answer:

    10 <-> 20 <-> 30 <-> null -> Option D
  4. Quick Check:

    New node at front = 10 first [OK]
Hint: New node points to old head; update old head's prev [OK]
Common Mistakes:
  • Not updating old head's prev pointer
  • Appending new node at end instead of beginning
  • Not updating head to new node
4. Identify the error in this code snippet for inserting at the beginning of a doubly linked list:
Node newNode = new Node(5);
newNode.next = head;
head.prev = newNode;
head = newNode;
medium
A. Incorrect assignment of newNode.next
B. Missing check if head is null before accessing head.prev
C. head should be assigned before setting newNode.next
D. newNode.prev should be set to head

Solution

  1. Step 1: Analyze head.prev assignment

    Code sets head.prev = newNode without checking if head is null, causing error if list is empty.
  2. Step 2: Identify missing null check

    Must check if head != null before setting head.prev to avoid null reference error.
  3. Final Answer:

    Missing check if head is null before accessing head.prev -> Option B
  4. Quick Check:

    Check head != null before head.prev [OK]
Hint: Always check head != null before accessing head.prev [OK]
Common Mistakes:
  • Ignoring null check on head
  • Setting pointers in wrong order
  • Confusing prev and next pointers
5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25 <-> null. You want to insert a new node with value 1 at the beginning. After insertion, which of the following correctly represents the list and the pointers of the new node?
hard
A. 5 <-> 15 <-> 25 <-> 1 <-> null, newNode.prev = 25, newNode.next = null
B. 1 <-> 5 <-> 15 <-> 25 <-> null, newNode.prev = 5, newNode.next = null
C. 1 <-> 5 <-> 15 <-> 25 <-> null, newNode.prev = null, newNode.next = 5
D. 5 <-> 15 <-> 25 <-> 1 <-> null, newNode.prev = null, newNode.next = 5

Solution

  1. Step 1: Insert new node at beginning

    New node with value 1 becomes head, points next to old head (5), prev is null.
  2. Step 2: Confirm pointers and list order

    List order is 1 <-> 5 <-> 15 <-> 25 <-> null, newNode.prev = null, newNode.next = 5.
  3. Final Answer:

    1 <-> 5 <-> 15 <-> 25 <-> null, newNode.prev = null, newNode.next = 5 -> Option C
  4. Quick Check:

    New head prev=null, next=old head [OK]
Hint: New head prev=null, next=old head node [OK]
Common Mistakes:
  • Setting newNode.prev to old head
  • Appending new node at end instead of beginning
  • Not updating head pointer
Recommended

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10 questions~5 mineasy hard