Merge Two Sorted Linked Lists in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

ListNode newNode = new ListNode([1]);

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

Ahead

Bval

Cnode

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using variable names that are not defined like 'value' or 'node'.

Confusing the head node with the 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 Java

current = current.[1];

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

Anext

Bchild

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which points backward.

Using undefined pointers like 'child' or 'head'.

To move forward in a linked list, we use the 'next' pointer.

3fill in blank

hard

Fix the error in the condition to check if the first list is empty.

DSA Java

if (list1 [1] null) { return list2; }

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

A==

B!=

C=

D=>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which assigns instead of compares.

Using '!=' which checks for not null.

To check equality in Java, use '==', not '=' which is assignment.

4fill in blank

hard

Fill both blanks to correctly compare node values and assign the next pointer.

DSA Java

if (list1.val [1] list2.val) {
    merged.next = list1;
    list1 = list1.[2];
}

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

A<

Bnext

C>

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' which reverses the order.

Using 'prev' which points backward.

We compare if list1's value is less than list2's, then move list1 to its next node.

5fill in blank

hard

Fill all three blanks to complete the while loop that merges two sorted lists.

DSA Java

while (list1 != null && list2 != null) {
    if (list1.val [1] list2.val) {
        merged.next = list1;
        list1 = list1.[2];
    } else {
        merged.next = list2;
        list2 = list2.[3];
    }
    merged = merged.next;
}

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

A<=

Bnext

Cprev

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' which reverses the logic.

Using 'prev' which moves backward.

We use '<=' to compare values, and 'next' to move pointers forward in both lists.

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 both linked lists

C. To find the maximum element in both lists

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the input lists

   Two linked lists are sorted individually, meaning their elements are in ascending order.

2. Step 2: Merge while maintaining order

   The goal is to combine these two lists into one sorted list that contains all elements from both.

3. Final Answer:

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

4. Quick Check:

   Merging sorted lists means combining all elements in order [OK]

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

Common Mistakes:

• Thinking merge means reversing lists
• Assuming merge removes duplicates
• Confusing merge with finding max element

2. Which of the following Java method signatures correctly represents a function to merge two sorted linked lists of integers?

easy

A. public int merge(ListNode l1, ListNode l2)

B. public void merge(int l1, int l2)

C. public ListNode merge(ListNode l1, ListNode l2)

D. public ListNode merge(int[] l1, int[] l2)

Solution

1. Step 1: Identify correct parameter types

   The function should accept two linked lists, so parameters must be ListNode types, not primitives or arrays.

2. Step 2: Identify correct return type

   The function returns a merged linked list, so return type should be ListNode, not void or int.

3. Final Answer:

   public ListNode merge(ListNode l1, ListNode l2) -> Option C

4. Quick Check:

   Method with ListNode params and return type is correct [OK]

Hint: Method merges two ListNode inputs and returns ListNode [OK]

Common Mistakes:

• Using primitive types instead of ListNode
• Returning void instead of ListNode
• Using arrays instead of linked lists

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

List 1: 1 -> 3 -> 5 -> null
List 2: 2 -> 4 -> 6 -> null

medium

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

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

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

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

Solution

1. Step 1: Compare first nodes of both lists

   1 (List 1) is less than 2 (List 2), so 1 is first in merged list.

2. Step 2: Continue comparing next nodes

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

3. Final Answer:

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

4. Quick Check:

   Merge keeps ascending order combining both lists [OK]

Hint: Pick smaller head node repeatedly to merge sorted lists [OK]

Common Mistakes:

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

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

public ListNode merge(ListNode l1, ListNode 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. Missing base case for both lists being null

B. No error; code correctly merges two sorted linked lists

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

D. Function does not handle empty lists properly

Solution

1. Step 1: Check base cases

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

2. Step 2: Check recursive merge logic

   Compares values, recursively merges next nodes, and returns the smaller node, which is correct.

3. Final Answer:

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

4. Quick Check:

   Recursive merge with base cases is correct [OK]

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

Common Mistakes:

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

5. You have two sorted linked lists:

List 1: 1 -> 4 -> 5 -> null
List 2: 1 -> 3 -> 4 -> null

After merging, what is the correct merged list?
Also, explain why the merged list contains duplicate values.

hard

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

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

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

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

Solution

1. Step 1: Merge nodes in ascending order

   Compare heads: both 1, add both 1s. Then 4 vs 3, add 3. Then 4 vs 4, add both 4s. Finally add 5.

2. Step 2: Understand duplicates in merged list

   Duplicates appear because both lists contain the same values independently; merge keeps all elements without removing duplicates.

3. Final Answer:

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

4. Quick Check:

   Merge keeps all elements including duplicates [OK]

Hint: Merge keeps duplicates if present in input lists [OK]

Common Mistakes:

• Removing duplicates during merge
• Appending lists without sorting
• Skipping nodes with same values