Merge Two Sorted Linked Lists in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = [1]

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

Aval

B0

CNone

Dself

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to 0 or val instead of None.

Confusing next with the node's value.

When creating a new node, its next pointer should initially be None because it doesn't point to any other node yet.

2fill in blank

medium

Complete the code to start merging by creating a dummy node.

DSA Python

def mergeTwoLists(l1, l2):
    dummy = Node([1])
    current = dummy
    # merging logic follows

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

A-1

BNone

Cl1.val

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using None as a value which causes errors.

Using l1.val which may not exist if l1 is None.

A dummy node with a placeholder value like -1 helps simplify the merge logic without affecting the final list.

3fill in blank

hard

Fix the error in the while loop condition to merge until one list ends.

DSA Python

while l1 [1] None and l2 != None:
    if l1.val < 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 stops the loop immediately.

Using comparison operators like '<' or '>' which are invalid here.

The loop should continue while both l1 and l2 are not None, so the condition uses '!=' to check they exist.

4fill in blank

hard

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

DSA Python

if l1 [1] None:
    current.next = l1
else:
    current.next = [2]

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

Ais not

Bis

Cl2

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'is' instead of 'is not' in the if condition.

Attaching None instead of the remaining list.

If l1 still has nodes (is not None), attach it; otherwise attach l2.

5fill in blank

hard

Fill the blanks to return the merged list skipping the dummy node.

DSA Python

def mergeTwoLists(l1, l2):
    dummy = Node(-1)
    current = dummy
    while l1 != None and l2 != None:
        if l1.val < l2.val:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next
        current = current.next
    if l1 is not None:
        current.next = l1
    else:
        current.next = l2
    return [1].[2]

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

Adummy

Bnext

Chead

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Returning dummy instead of dummy.next.

Returning current or head which are not defined here.

Return dummy.next to skip the dummy node and get the head of the merged list.

Practice

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

easy

A. To create a new sorted linked list containing all elements from both lists

B. To reverse the order of elements in both lists

C. To find the largest element in both lists

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the merging process

   Merging two sorted linked lists means combining them into one list that keeps the sorted order.

2. Step 2: Identify the goal

   The goal is to have a new linked list that contains all elements from both lists in sorted order.

3. Final Answer:

   To create a new sorted linked list containing all elements from both lists -> Option A

4. Quick Check:

   Merging = combine sorted lists [OK]

Hint: Merging keeps order and combines all elements [OK]

Common Mistakes:

• Thinking merging reverses lists
• Only finding largest element
• Removing duplicates instead of merging

2. Which of the following is the correct way to start merging two sorted linked lists in Python?

easy

A. def mergeLists(l1, l2):
current = ListNode(0)
dummy = current.next

B. def mergeLists(l1, l2):
dummy = null
current = dummy

C. def mergeLists(l1, l2):
dummy = ListNode(0)
current = dummy

D. def mergeLists(l1, l2):
dummy = ListNode(0)
current = dummy.next

Solution

1. Step 1: Initialize dummy and current nodes properly

   We create a dummy node with value 0 and set current to dummy to build the merged list.

2. Step 2: Check options for correct syntax

   def mergeLists(l1, l2): dummy = ListNode(0) current = dummy correctly creates dummy and current pointing to dummy. def mergeLists(l1, l2): dummy = ListNode(0) current = dummy.next incorrectly sets current = dummy.next which is initially None.

3. Final Answer:

   def mergeLists(l1, l2): dummy = ListNode(0) current = dummy -> Option C

4. Quick Check:

   Dummy node with value 0 starts merge [OK]

Hint: Start with dummy node value 0 and current pointing to dummy [OK]

Common Mistakes:

• Setting current to dummy.next
• Swapping dummy and current incorrectly
• Setting dummy to null

3. What will be the output linked list after merging these two sorted lists? List 1: 1 -> 3 -> 5 -> null List 2: 2 -> 4 -> 6 -> null Use the merge logic to combine them.

medium

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

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

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

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

Solution

1. Step 1: Compare nodes from both lists in order

   Start with 1 (List 1) and 2 (List 2). 1 is smaller, add 1. Then compare 3 and 2, add 2. Continue comparing and adding smaller nodes.

2. Step 2: Build merged list step-by-step

   Merged list becomes 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null after adding all nodes in sorted order.

3. Final Answer:

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

4. Quick Check:

   Sorted merge combines nodes in ascending order [OK]

Hint: Pick smaller node each step to keep order [OK]

Common Mistakes:

• Appending one list fully before the other
• Mixing node order incorrectly
• Skipping nodes during merge

4. Identify the error in this merge function snippet:

def mergeLists(l1, l2):
    dummy = ListNode(0)
    current = dummy
    while l1 and l2:
        if l1.val < l2.val:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next
    current.next = l1 if l1 else l2
    return dummy.next

medium

A. The 'current' pointer is not moved forward inside the loop

B. The dummy node is not initialized correctly

C. The function does not handle empty input lists

D. The return statement should return 'current' instead of 'dummy.next'

Solution

1. Step 1: Check pointer updates inside the loop

   After assigning current.next, the current pointer must move forward to current.next to continue building the list.

2. Step 2: Identify missing current update

   The code misses 'current = current.next' inside the while loop, causing an infinite loop or incorrect list.

3. Final Answer:

   The 'current' pointer is not moved forward inside the loop -> Option A

4. Quick Check:

   Move current forward after linking nodes [OK]

Hint: Always move current pointer after linking a node [OK]

Common Mistakes:

• Forgetting to move current pointer
• Returning wrong node
• Not handling empty lists (though not error here)

5. Given two sorted linked lists where one list is empty and the other is 0 -> 2 -> 4 -> null, what should the merged list be?

medium

A. 0 -> null

B. null

C. 4 -> 2 -> 0 -> null

D. 0 -> 2 -> 4 -> null

Solution

1. Step 1: Understand merging with an empty list

   If one list is empty, the merged list is just the non-empty list as it is already sorted.

2. Step 2: Apply to given lists

   Since the second list is empty and the first is 0 -> 2 -> 4 -> null, the merged list remains 0 -> 2 -> 4 -> null.

3. Final Answer:

   0 -> 2 -> 4 -> null -> Option D

4. Quick Check:

   Merging with empty list returns the other list [OK]

Hint: Empty list merge returns the other list unchanged [OK]

Common Mistakes:

• Returning null when one list is empty
• Reversing the non-empty list
• Returning only the first node