Add Two Numbers Represented as Linked List in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the sum digit.

DSA Java

ListNode newNode = new ListNode([1]);

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

Asum % 10

Bsum + 10

Csum / 10

Dsum * 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum / 10 instead of sum % 10.

Adding 10 to sum instead of modulo operation.

We use sum % 10 to get the digit to store in the new node because sum can be more than 9.

2fill in blank

medium

Complete the code to update the carry after adding two digits.

DSA Java

carry = [1];

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

Asum % 10

Bsum + 10

Csum * 10

Dsum / 10

Attempts:

3 left

💡 Hint

Common Mistakes

Using sum % 10 for carry instead of sum / 10.

Adding 10 to sum instead of dividing.

Carry is the tens digit of the sum, so we use integer division sum / 10.

3fill in blank

hard

Fix the error in the loop condition to continue while nodes or carry exist.

DSA Java

while (l1 != null || [1] != null || carry != 0) {

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

Acarry

Bl2

Cl1

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Checking carry twice instead of l2.

Checking l1 twice instead of l2.

We must check if l2 is not null to continue adding digits from the second list.

4fill in blank

hard

Fill both blanks to safely get values from nodes or zero if null.

DSA Java

int x = (l1 != null) ? [1] : 0;
int y = (l2 != null) ? [2] : 0;

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Al1.val

Bl2.val

Cl1.next

Dl2.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using .next instead of .val to get values.

Not checking for null before accessing values.

We get the value from l1.val and l2.val if nodes exist; otherwise zero.

5fill in blank

hard

Fill all three blanks to move to the next nodes or keep null if none.

DSA Java

l1 = (l1 != null) ? [1] : null;
l2 = (l2 != null) ? [2] : null;
current = current.[3];

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

Al1.next

Bl2.next

Cnext

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using .val instead of .next to move nodes.

Not updating current pointer correctly.

We move l1 and l2 to their next nodes, and current to current.next to continue building the list.

Practice

1. What does the problem "Add Two Numbers Represented as Linked List" mainly involve?

easy

A. Reversing a linked list

B. Sorting two linked lists in ascending order

C. Finding the middle node of a linked list

D. Adding digits node by node from two linked lists representing numbers

Solution

1. Step 1: Understand the problem context

   The problem involves two linked lists where each node contains a digit of a number, stored in reverse order.

2. Step 2: Identify the required operation

   We need to add these two numbers digit by digit, handling carry, and return the sum as a linked list.

3. Final Answer:

   Adding digits node by node from two linked lists representing numbers -> Option D

4. Quick Check:

   Adding digits node by node = D [OK]

Hint: Focus on digit-wise addition with carry [OK]

Common Mistakes:

• Confusing addition with sorting
• Ignoring carry during addition
• Assuming numbers are stored in forward order

2. Which of the following Java code snippets correctly initializes a new node for a singly linked list representing a digit?

easy

A. ListNode node = ListNode(5);

B. ListNode node = new ListNode(); node.next = 5;

C. ListNode node = new ListNode(5);

D. ListNode node = new ListNode.val(5);

Solution

1. Step 1: Recall Java object creation syntax

   In Java, to create a new object, use the 'new' keyword followed by the constructor call with parameters.

2. Step 2: Check each option for correct syntax

   ListNode node = new ListNode(5); uses 'new ListNode(5);' which is correct. ListNode node = ListNode(5); misses 'new'. ListNode node = new ListNode(); node.next = 5; incorrectly sets next instead of val. ListNode node = new ListNode.val(5); is invalid syntax.

3. Final Answer:

   ListNode node = new ListNode(5); -> Option C

4. Quick Check:

   Java new object syntax = B [OK]

Hint: Use 'new' keyword with constructor for object creation [OK]

Common Mistakes:

• Omitting 'new' keyword
• Incorrect constructor call syntax
• Trying to call constructor like a method without 'new'
• Confusing node.val and node.next

3. What is the output linked list after adding the two numbers represented by these linked lists? List 1: 2 -> 4 -> 3 -> null (represents 342) List 2: 5 -> 6 -> 4 -> null (represents 465)

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummy;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummy.next;
}

medium

A. 7 -> 0 -> 8 -> null

B. 8 -> 0 -> 7 -> null

C. 7 -> 10 -> 7 -> null

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

Solution

1. Step 1: Add digits node by node with carry

   Adding 2+5=7 (carry 0), 4+6=10 (carry 1), 3+4=7 plus carry 1 = 8.

2. Step 2: Construct the resulting linked list

   Digits are 7 (units), 0 (tens), 8 (hundreds), so linked list is 7 -> 0 -> 8 -> null.

3. Final Answer:

   7 -> 0 -> 8 -> null -> Option A

4. Quick Check:

   342 + 465 = 807 represented as 7->0->8 [OK]

Hint: Add digits with carry, build new nodes [OK]

Common Mistakes:

• Ignoring carry in addition
• Appending nodes incorrectly
• Mixing digit order in output

4. Identify the bug in this code snippet for adding two numbers represented as linked lists:

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode curr = dummy;
    int carry = 0;
    while (l1 != null && l2 != null) {
        int sum = carry + l1.val + l2.val;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        l1 = l1.next;
        l2 = l2.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummy.next;
}

medium

A. The carry is not calculated correctly

B. The while loop should use '||' instead of '&&' to handle different length lists

C. The dummy node is not initialized properly

D. The function does not return the correct node

Solution

1. Step 1: Analyze the loop condition

   The loop uses '&&' which stops when either list ends, missing remaining nodes of longer list.

2. Step 2: Correct loop condition to '||'

   Using '||' ensures the loop continues until both lists are fully processed, handling different lengths correctly.

3. Final Answer:

   The while loop should use '||' instead of '&&' to handle different length lists -> Option B

4. Quick Check:

   Loop condition must cover all nodes = A [OK]

Hint: Use '||' to process all nodes in both lists [OK]

Common Mistakes:

• Using '&&' causing early loop exit
• Not handling carry after loop
• Not advancing pointers correctly

5. Given two linked lists representing numbers in reverse order, how would you modify the addition algorithm to handle the case where the input lists are extremely long (millions of nodes) and memory is limited?

hard

A. Use iterative addition with constant extra space and process nodes one by one

B. Convert linked lists to integers, add, then convert back to linked list

C. Store all digits in arrays before adding to optimize speed

D. Reverse both lists first, add, then reverse the result

Solution

1. Step 1: Consider memory constraints with large inputs

   Converting to integers or arrays requires large memory, not feasible for millions of nodes.

2. Step 2: Use iterative addition with constant extra space

   Process nodes one by one, create result nodes on the fly, and avoid storing entire numbers in memory.

3. Final Answer:

   Use iterative addition with constant extra space and process nodes one by one -> Option A

4. Quick Check:

   Iterative node-wise addition saves memory = C [OK]

Hint: Add nodes one by one without full number conversion [OK]

Common Mistakes:

• Trying to convert entire list to integer causing overflow
• Using extra arrays increasing memory usage
• Reversing lists unnecessarily increasing complexity