Merge Two Sorted Linked Lists in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function ListNode(val) {
  this.val = val;
  this.next = [1];
}

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

A0

Bundefined

Cfalse

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined or 0 instead of null for the next pointer.

The next pointer of a new node should be null to indicate the end of the list.

2fill in blank

medium

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

DSA Javascript

let dummy = new ListNode([1]);
let current = dummy;

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

Anull

B-1

Cundefined

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using null or undefined as node value causes errors.

Using -1 as dummy node value is common to avoid confusion with list values.

3fill in blank

hard

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

DSA Javascript

while (list1 !== null && list2 !== null) {
  if (list1.val [1] list2.val) {
    current.next = list1;
    list1 = list1.next;
  } else {
    current.next = list2;
    list2 = list2.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 '>' picks larger values first, breaking sorted order.

We pick the smaller value node first, so use less than operator.

4fill in blank

hard

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

DSA Javascript

if (list1 !== null) {
  current.next = [1];
} else {
  current.next = [2];
}

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

Alist1

Blist2

Cdummy

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Attaching dummy or current instead of the remaining list.

Attach whichever list still has nodes left to the merged list.

5fill in blank

hard

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

DSA Javascript

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 or current directly instead of dummy.next.

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

Practice

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

easy

A. To find the largest element in both lists

B. To reverse both linked lists

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

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the merging process

   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 lists without changing their order.

3. Final Answer:

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

4. Quick Check:

   Merging = combining sorted lists [OK]

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

Common Mistakes:

• Thinking merging reverses the lists
• Confusing merging with finding max element
• Assuming duplicates are removed automatically

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

easy

A. let dummy = new ListNode(0); let current = dummy;

B. let dummy = new ListNode(); let current = dummy.next;

C. let dummy = 0; let current = dummy;

D. let current = null; let dummy = current;

Solution

1. Step 1: Initialize a dummy node correctly

   We create a dummy node with a value (usually 0) to start the merged list.

2. Step 2: Set a pointer to track the current node

   We set current to dummy to build the merged list from dummy onwards.

3. Final Answer:

   let dummy = new ListNode(0); let current = dummy; -> Option A

4. Quick Check:

   Dummy node initialization = let dummy = new ListNode(0); let current = dummy; [OK]

Hint: Start with dummy node and current pointer for easy list building [OK]

Common Mistakes:

• Using primitive 0 instead of ListNode object
• Swapping dummy and current incorrectly
• Setting current or dummy to null initially

3. What will be the merged linked list after running the following code?

function ListNode(val, next = null) { this.val = val; this.next = next; }

let l1 = new ListNode(1, new ListNode(3, new ListNode(5)));
let l2 = new ListNode(2, new ListNode(4, new ListNode(6)));

function mergeTwoLists(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  while (l1 && l2) {
    if (l1.val < l2.val) {
      current.next = l1;
      l1 = l1.next;
    } else {
      current.next = l2;
      l2 = l2.next;
    }
    current = current.next;
  }
  current.next = l1 || l2;
  return dummy.next;
}

let merged = mergeTwoLists(l1, l2);

let result = [];
while (merged) {
  result.push(merged.val);
  merged = merged.next;
}
console.log(result);

medium

A. [1, 3, 5, 2, 4, 6]

B. [1, 2, 3, 4, 5, 6]

C. [2, 4, 6, 1, 3, 5]

D. [6, 5, 4, 3, 2, 1]

Solution

1. Step 1: Trace the merging loop

   Compare nodes: 1 < 2, add 1; 3 > 2, add 2; 3 < 4, add 3; 5 > 4, add 4; 5 < 6, add 5; then add remaining 6.

2. Step 2: Collect merged list values

   The merged list values in order are [1, 2, 3, 4, 5, 6].

3. Final Answer:

   [1, 2, 3, 4, 5, 6] -> Option B

4. Quick Check:

   Sorted merge result = [1, 2, 3, 4, 5, 6] [OK]

Hint: Compare heads, pick smaller, move pointer, repeat [OK]

Common Mistakes:

• Appending entire list instead of node by node
• Not updating current pointer after adding node
• Mixing order of nodes in merged list

4. Identify the bug in this merge function and select the fix:

function mergeTwoLists(l1, l2) {
  let dummy = new ListNode(0);
  let current = dummy;
  while (l1 !== null || l2 !== null) {
    if (l1.val < l2.val) {
      current.next = l1;
      l1 = l1.next;
    } else {
      current.next = l2;
      l2 = l2.next;
    }
    current = current.next;
  }
  return dummy.next;
}

medium

A. Initialize dummy with null instead of 0

B. Add check for null before comparing l1.val and l2.val

C. Remove current = current.next inside the loop

D. Change while condition to 'while (l1 && l2)'

Solution

1. Step 1: Analyze the while loop condition

   The loop uses '||' which allows l1 or l2 to be null, causing errors when accessing val.

2. Step 2: Fix the loop condition

   Changing the condition to 'while (l1 && l2)' ensures both lists have nodes before accessing val.

3. Final Answer:

   Change while condition to 'while (l1 && l2)' -> Option D

4. Quick Check:

   Loop condition with && prevents null pointer errors [OK]

Hint: Use && in loop condition to ensure both nodes exist before access [OK]

Common Mistakes:

• Using || instead of && in loop condition without null checks
• Forgetting to move current pointer
• Initializing dummy with wrong value type

5. Given two sorted linked lists where one is empty and the other is 1 -> 2 -> 3 -> null, what is the merged list?

medium

A. 1 -> 2 -> 3 -> null

B. Error due to empty list

C. 3 -> 2 -> 1 -> null

D. null

Solution

1. Step 1: Understand merging with an empty list

   If one list is empty, the merged list is just the other list unchanged.

2. Step 2: Apply this to given lists

   Since one list is empty and the other is 1 -> 2 -> 3 -> null, the merged list is 1 -> 2 -> 3 -> null.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option A

4. Quick Check:

   Merging with empty list returns the non-empty list [OK]

Hint: Empty list merged with sorted list returns the sorted list [OK]

Common Mistakes:

• Assuming error occurs when one list is empty
• Reversing the non-empty list mistakenly
• Returning null instead of the non-empty list