Merge Two Sorted Linked Lists in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

ListNode newNode = new ListNode([1]);

Drag options to blanks, or click blank then click option'

Anext

Bnode

Cval

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'next' instead of the value.

Using 'head' which is a node, not a value.

The constructor of ListNode takes the value 'val' to create a new node.

2fill in blank

medium

Complete the code to move the pointer to the next node in the list.

DSA C#

current = current[1];

Drag options to blanks, or click blank then click option'

Aprev

Bnext

Cval

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which points backward.

Using 'val' which is the node's value, not a pointer.

To move forward in a linked list, we assign current to current.next.

3fill in blank

hard

Fix the error in the if condition to compare node values correctly.

DSA C#

if (list1[1].val <= list2.val) {

Drag options to blanks, or click blank then click option'

Anext

Bprev

Chead

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing list1 directly without accessing its next node.

Using 'prev' or 'head' which are not valid here.

We must check the value of the next node in list1 to compare with list2's current value.

4fill in blank

hard

Fill both blanks to correctly attach the smaller node and move the pointer.

DSA C#

current.next = [1];
current = [2];

Drag options to blanks, or click blank then click option'

Alist1

Blist2

Ccurrent.next

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Attaching list2 when list1 is smaller.

Moving current to 'current' instead of 'current.next'.

We attach list1 to current.next and then move current to current.next to continue.

5fill in blank

hard

Fill all three blanks to return the merged list starting after the dummy node.

DSA C#

return [1].[2];

// Also, update pointers:
list1 = list1.[3];

Drag options to blanks, or click blank then click option'

Adummy

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dummy instead of dummy.next.

Using 'prev' instead of 'next' to move pointers.

Return dummy.next as the head of merged list and move list1 to list1.next.

Practice

1. What is the main goal when merging two sorted linked lists?

easy

A. Remove duplicates from both lists

B. Create a new sorted linked list containing all elements from both lists

C. Reverse both linked lists

D. Sort each list individually without merging

Solution

1. Step 1: Understand the problem goal

   The goal is to combine two sorted linked lists into one sorted linked list.

2. Step 2: Clarify the expected output

   The output should be a single linked list that maintains sorted order and contains all nodes from both input lists.

3. Final Answer:

   Create a new sorted linked list containing all elements from both lists -> Option B

4. Quick Check:

   Merge means combine sorted lists into one sorted list [OK]

Hint: Merging means combining sorted lists into one sorted list [OK]

Common Mistakes:

• Thinking merge means reversing lists
• Trying to sort lists separately without merging
• Removing duplicates instead of merging

2. Which of the following is the correct C# syntax to declare a node class for a singly linked list used in merging?

easy

A. class Node { int val; Node next; Node(int x) { val = x; next = null; } }

B. public struct Node { int val; Node next; }

C. public class Node { public int val; public Node next; public Node(int x) { val = x; next = null; } }

D. public class Node { int val; int next; public Node(int x) { val = x; next = 0; } }

Solution

1. Step 1: Check class declaration and access modifiers

   public class Node { public int val; public Node next; public Node(int x) { val = x; next = null; } } correctly declares a public class with public fields and a constructor initializing values.

2. Step 2: Verify field types and initialization

   public class Node { public int val; public Node next; public Node(int x) { val = x; next = null; } } uses 'int val' and 'Node next' with proper initialization; others miss public or use wrong types.

3. Final Answer:

   public class Node { public int val; public Node next; public Node(int x) { val = x; next = null; } } -> Option C

4. Quick Check:

   Correct class with public fields and constructor [OK]

Hint: Node class needs public fields and constructor for linked list [OK]

Common Mistakes:

• Missing public keyword causing access errors
• Using struct instead of class for linked list node
• Incorrect type for next pointer

3. What will be the output linked list after merging these two sorted lists?

List1: 1 -> 3 -> 5 -> null
List2: 2 -> 4 -> 6 -> null

Assuming the merge function merges correctly.

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> null

C. 2 -> 4 -> 6 -> 1 -> 3 -> 5 -> null

D. 1 -> 2 -> 4 -> 3 -> 5 -> 6 -> null

Solution

1. Step 1: Compare first nodes of both lists

   1 (List1) < 2 (List2), so 1 is first in merged list.

2. Step 2: Continue comparing nodes in order

   Next compare 3 and 2, 2 is smaller, then 3 and 4, 3 is smaller, then 5 and 4, 4 is smaller, then 5 and 6, 5 is smaller, finally 6.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null -> Option A

4. Quick Check:

   Merge sorted lists by picking smaller node each step [OK]

Hint: Pick smaller node from heads repeatedly to merge [OK]

Common Mistakes:

• Appending one list entirely before the other
• Not maintaining sorted order
• Mixing node order incorrectly

4. Given the following code snippet for merging two sorted linked lists, what is the error?

public Node Merge(Node l1, Node l2) {
    if (l1 == null) return l2;
    if (l2 == null) return l1;
    if (l1.val < l2.val) {
        l1.next = Merge(l1.next, l2);
        return l1;
    } else {
        l2.next = Merge(l1, l2.next);
        return l2;
    }
}

medium

A. No error; the code correctly merges two sorted linked lists

B. Missing base case for empty lists

C. Incorrect comparison operator used

D. The function does not handle cycles in lists

Solution

1. Step 1: Analyze base cases

   Code returns the other list if one is null, correctly handling empty lists.

2. Step 2: Check recursive merge logic

   Code compares values and recursively merges the rest, correctly linking nodes in sorted order.

3. Final Answer:

   No error; the code correctly merges two sorted linked lists -> Option A

4. Quick Check:

   Recursive merge with base cases and correct comparisons [OK]

Hint: Check base cases and recursive calls carefully [OK]

Common Mistakes:

• Forgetting to return merged list
• Incorrect base case handling
• Wrong comparison operator

5. You have two sorted linked lists:
List1: 1 -> 4 -> 5 -> null
List2: 1 -> 3 -> 4 -> null

After merging, what is the correct merged list?
Also, explain why duplicates appear and how the merge handles them.

hard

A. 1 -> 1 -> 4 -> 3 -> 4 -> 5 -> null

B. 1 -> 3 -> 4 -> 5 -> null

C. 1 -> 4 -> 5 -> 3 -> 4 -> 1 -> null

D. 1 -> 1 -> 3 -> 4 -> 4 -> 5 -> null

Solution

1. Step 1: Merge nodes in sorted order including duplicates

   Both lists have '1' and '4' values; merge keeps all nodes, so duplicates appear in order.

2. Step 2: Explain duplicate handling

   Merge does not remove duplicates; it links nodes as they appear, preserving duplicates in sorted order.

3. Final Answer:

   1 -> 1 -> 3 -> 4 -> 4 -> 5 -> null -> Option D

4. Quick Check:

   Duplicates kept; merge preserves all nodes sorted [OK]

Hint: Merge keeps duplicates; just link nodes in sorted order [OK]

Common Mistakes:

• Removing duplicates mistakenly
• Mixing node order
• Appending lists without sorting