Merge Two Sorted Linked Lists in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

newNode := &ListNode{Val: [1], Next: nil}

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

Aval

Bvalue

C5

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name instead of a number.

Using incorrect field names.

We create a new node with the value 5. The field name is Val, so we assign 5 to Val.

2fill in blank

medium

Complete the code to start merging from a dummy node.

DSA Go

dummy := &ListNode{}
current := [1]

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

Adummy

Bcurrent

Chead

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning current to nil or head instead of dummy.

Using an undefined variable.

We start merging from the dummy node, so current points to dummy.

3fill in blank

hard

Fix the error in the condition to compare node values.

DSA Go

for l1 != nil && l2 != nil {
    if l1.Val [1] l2.Val {
        current.Next = l1
        l1 = l1.Next
    } else {
        current.Next = l2
        l2 = l2.Next
    }
    current = current.Next
}

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

A<

B>

C==

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which picks the larger value first.

Using '==' or '!=' which do not order nodes.

We want to pick the smaller value node first, so we use '<' to compare.

4fill in blank

hard

Fill both blanks to attach the remaining nodes after the loop.

DSA Go

if l1 != nil {
    current.Next = [1]
} else {
    current.Next = [2]
}

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

Al1

Bl2

Ccurrent

Ddummy

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning current or dummy instead of l1 or l2.

Swapping l1 and l2 in the else block.

After the loop, attach the remaining nodes from l1 or l2 to current.Next.

5fill in blank

hard

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

DSA Go

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
    dummy := &ListNode{}
    current := dummy
    for l1 != nil && l2 != nil {
        if l1.Val < l2.Val {
            current.Next = l1
            l1 = l1.Next
        } else {
            current.Next = l2
            l2 = l2.Next
        }
        current = current.Next
    }
    if l1 != nil {
        current.Next = l1
    } else {
        current.Next = l2
    }
    return [1]
}

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

Adummy

Bcurrent

Ccurrent.Next

Ddummy.Next

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dummy instead of dummy.Next.

Returning current or current.Next which may not be the head.

We return dummy.Next because dummy is a placeholder node before the merged list.

Practice

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

easy

A. Remove duplicates from both lists without merging

B. Sort each list separately without combining them

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

D. Reverse both lists and then merge

Solution

1. Step 1: Understand the merge operation

   Merging means combining two lists into one while keeping the order sorted.

2. Step 2: Identify the goal of merging sorted lists

   The goal is to create a new sorted list that contains all elements from both input lists.

3. Final Answer:

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

4. Quick Check:

   Merge = Combine sorted lists [OK]

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

Common Mistakes:

• Thinking merge means sorting each list separately
• Confusing merge with removing duplicates only
• Assuming merge reverses the lists

2. Which of the following is the correct Go syntax to define a node struct for a singly linked list?

easy

A. Node := struct { val int; next Node }

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

C. class Node { int val; Node next; }

D. type Node struct { val int; next *Node }

Solution

1. Step 1: Recall Go struct syntax

   In Go, structs are defined with 'type Name struct { fields }' and pointers use '*'.

2. Step 2: Check each option

   type Node struct { val int; next *Node } uses correct Go syntax with pointer '*Node' for next node.

3. Final Answer:

   type Node struct { val int; next *Node } -> Option D

4. Quick Check:

   Go struct with pointer = type Node struct { val int; next *Node } [OK]

Hint: Go structs use 'type' and pointers with '*' [OK]

Common Mistakes:

• Using C/C++ or Java syntax in Go
• Missing pointer '*' for next node
• Using := inside struct definition

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

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

Assume the merge function returns the head of the merged list.

medium

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

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

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

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

Solution

1. Step 1: Compare first nodes of both lists

   1 (List1) < 2 (List2), so start merged list with 1.

2. Step 2: Continue comparing next nodes in order

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

3. Final Answer:

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

4. Quick Check:

   Merge sorted lists = sorted combined list [OK]

Hint: Pick smaller node each step to merge sorted lists [OK]

Common Mistakes:

• Appending one list after the other without sorting
• Reversing the order accidentally
• Skipping nodes during merge

4. Given this Go function snippet to merge two sorted linked lists, what is the bug?

func mergeLists(l1, l2 *Node) *Node {
    if l1 == nil {
        return l2
    }
    if l2 == nil {
        return l1
    }
    if l1.val < l2.val {
        l1.next = mergeLists(l1.next, l2)
        return l1
    } else {
        l2.next = mergeLists(l1, l2.next)
        return l2
    }
}

medium

A. No bug, function correctly merges two sorted lists

B. Missing base case for both lists being nil

C. Incorrect comparison operator, should be <= instead of <

D. Function does not handle cycles in lists

Solution

1. Step 1: Analyze base cases

   Function returns the other list if one is nil, which is correct.

2. Step 2: Check recursive merge logic

   Function compares values and recursively merges the rest correctly.

3. Final Answer:

   No bug, function correctly merges two sorted lists -> Option A

4. Quick Check:

   Recursive merge with base cases = correct [OK]

Hint: Check base cases and recursion for correctness [OK]

Common Mistakes:

• Thinking missing base case for both nil (not needed)
• Confusing < with <= (both work here)
• Ignoring cycle detection (not required here)

5. You have two sorted linked lists:
List1: 1 -> 4 -> 7 -> 10 -> null
List2: 2 -> 3 -> 5 -> 8 -> 12 -> null

Write a Go function to merge these lists iteratively and return the merged list head. What will be the final merged list?

hard

A. 1 -> 2 -> 3 -> 4 -> 5 -> 7 -> 8 -> 10 -> 12 -> null

B. 1 -> 4 -> 7 -> 10 -> 2 -> 3 -> 5 -> 8 -> 12 -> null

C. 2 -> 3 -> 5 -> 8 -> 12 -> 1 -> 4 -> 7 -> 10 -> null

D. 12 -> 10 -> 8 -> 7 -> 5 -> 4 -> 3 -> 2 -> 1 -> null

Solution

1. Step 1: Initialize dummy node and pointer

   Create a dummy node to start merged list and a pointer to track current node.

2. Step 2: Iterate and compare nodes from both lists

   Compare current nodes from List1 and List2, attach smaller to merged list, move pointer forward.

3. Step 3: Attach remaining nodes after one list ends

   Once one list is exhausted, attach the rest of the other list.

4. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 7 -> 8 -> 10 -> 12 -> null -> Option A

5. Quick Check:

   Iterative merge keeps sorted order [OK]

Hint: Use dummy node and iterate picking smaller nodes [OK]

Common Mistakes:

• Appending lists without sorting
• Forgetting to attach remaining nodes
• Not using a dummy node causing complex code