Add Two Numbers Represented as Linked List in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Adding Two Numbers Represented by Linked Lists

What is the printed linked list after adding the two numbers represented by these linked lists?

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

The digits are stored in reverse order, each node contains a single digit.

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            int sum = carry;
            if (l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            carry = sum / 10;
            current.next = new ListNode(sum % 10);
            current = current.next;
        }
        return dummy.next;
    }

    public static void printList(ListNode node) {
        while (node != null) {
            System.out.print(node.val);
            if (node.next != null) System.out.print(" -> ");
            node = node.next;
        }
        System.out.println(" -> null");
    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(2);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(3);

        ListNode l2 = new ListNode(5);
        l2.next = new ListNode(6);
        l2.next.next = new ListNode(4);

        ListNode result = addTwoNumbers(l1, l2);
        printList(result);
    }
}

A 7 -> 0 -> 8 -> null

B 8 -> 0 -> 7 -> null

C 7 -> 0 -> 7 -> null

D 7 -> 1 -> 8 -> null

💡 Hint

Add digits node by node, remember to carry over when sum >= 10.

Adding 342 + 465 = 807. The linked list stores digits in reverse order: 7 -> 0 -> 8 -> null.

❓ Predict Output

intermediate

Result of Adding Unequal Length Linked Lists

What is the output linked list after adding these two numbers?

List 1: 9 -> 9 -> 9 -> 9 -> null
List 2: 1 -> null

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            int sum = carry;
            if (l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            carry = sum / 10;
            current.next = new ListNode(sum % 10);
            current = current.next;
        }
        return dummy.next;
    }

    public static void printList(ListNode node) {
        while (node != null) {
            System.out.print(node.val);
            if (node.next != null) System.out.print(" -> ");
            node = node.next;
        }
        System.out.println(" -> null");
    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(9);
        l1.next = new ListNode(9);
        l1.next.next = new ListNode(9);
        l1.next.next.next = new ListNode(9);

        ListNode l2 = new ListNode(1);

        ListNode result = addTwoNumbers(l1, l2);
        printList(result);
    }
}

A 0 -> 0 -> 0 -> 0 -> 1 -> null

B 0 -> 0 -> 0 -> 0 -> null

C 1 -> 0 -> 0 -> 0 -> 0 -> null

D 9 -> 9 -> 9 -> 10 -> null

💡 Hint

Remember to add carry even after both lists end.

Adding 9999 + 1 = 10000, stored as 0 -> 0 -> 0 -> 0 -> 1 -> null in reverse order.

🔧 Debug

advanced

Identify the Bug in Linked List Addition Code

What error will this code produce when adding two linked lists?

Code snippet:

while (l1 != null && l2 != null) {
    int sum = l1.val + l2.val + carry;
    carry = sum / 10;
    current.next = new ListNode(sum % 10);
    current = current.next;
    l1 = l1.next;
    l2 = l2.next;
}
// Missing code to handle remaining nodes or carry

A The code will produce an infinite loop because l1 and l2 are never updated.

B The code will miss adding remaining nodes if lists are unequal length, resulting in incomplete sum.

C The code will correctly add the numbers without any error.

D The code will cause a NullPointerException when accessing l1.val or l2.val.

💡 Hint

Check what happens after the while loop ends.

The loop only runs while both lists have nodes. If one list is longer, those nodes are ignored, so the sum is incomplete.

❓ Predict Output

advanced

Output When Adding Two Large Numbers with Multiple Carries

What is the output linked list after adding these two numbers?

List 1: 9 -> 8 -> 7 -> 6 -> 5 -> null
List 2: 1 -> 2 -> 3 -> 4 -> 5 -> null

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class Solution {
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            int sum = carry;
            if (l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            carry = sum / 10;
            current.next = new ListNode(sum % 10);
            current = current.next;
        }
        return dummy.next;
    }

    public static void printList(ListNode node) {
        while (node != null) {
            System.out.print(node.val);
            if (node.next != null) System.out.print(" -> ");
            node = node.next;
        }
        System.out.println(" -> null");
    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(9);
        l1.next = new ListNode(8);
        l1.next.next = new ListNode(7);
        l1.next.next.next = new ListNode(6);
        l1.next.next.next.next = new ListNode(5);

        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(2);
        l2.next.next = new ListNode(3);
        l2.next.next.next = new ListNode(4);
        l2.next.next.next.next = new ListNode(5);

        ListNode result = addTwoNumbers(l1, l2);
        printList(result);
    }
}

A 0 -> 1 -> 1 -> 1 -> 0 -> null

B 0 -> 1 -> 1 -> 1 -> 1 -> null

C 0 -> 1 -> 1 -> 1 -> 1 -> 1 -> null

D 0 -> 1 -> 1 -> 1 -> 0 -> 1 -> null

💡 Hint

Add digits carefully and track carry for each step.

Adding 56789 + 54321 = 111110, stored as 0 -> 1 -> 1 -> 1 -> 1 -> 1 -> null in reverse order.

🧠 Conceptual

expert

Why Use a Dummy Node in Adding Two Numbers as Linked Lists?

Why is a dummy node used at the start of the linked list when adding two numbers represented as linked lists?

A To keep track of the length of the resulting linked list.

B To store the carry value separately from the sum nodes.

C To reverse the linked list after addition is complete.

D To simplify code by avoiding special cases for the head node and easily return the result list.

💡 Hint

Think about how linked lists are built node by node.

The dummy node acts as a fixed starting point so we don't need to check if the head is null or not when adding nodes.

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