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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Adding Two Linked Lists Representing Numbers

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

List 1: 2 -> 4 -> 3 -> null (represents 342)
List 2: 5 -> 6 -> 4 -> null (represents 465)

Code adds the numbers and prints the resulting linked list.

DSA C

struct Node {
    int val;
    struct Node* next;
};

struct Node* addTwoNumbers(struct Node* l1, struct Node* l2) {
    struct Node* dummy = malloc(sizeof(struct Node));
    dummy->val = 0;
    dummy->next = NULL;
    struct Node* 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;
        struct Node* newNode = malloc(sizeof(struct Node));
        newNode->val = sum % 10;
        newNode->next = NULL;
        current->next = newNode;
        current = newNode;
    }
    struct Node* result = dummy->next;
    free(dummy);
    return result;
}

void printList(struct Node* head) {
    while (head != NULL) {
        printf("%d", head->val);
        if (head->next != NULL) printf(" -> ");
        head = head->next;
    }
    printf(" -> null\n");
}

// After calling addTwoNumbers with above lists, printList is called on result.

A 7 -> 0 -> 8 -> null

B 8 -> 0 -> 7 -> null

C 7 -> 10 -> 7 -> null

D 5 -> 10 -> 7 -> null

💡 Hint

Add digits from right to left, carry over if sum >= 10.

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

🧠 Conceptual

intermediate

Understanding Carry in Linked List Addition

Why is it necessary to continue the loop when carry is not zero after processing both linked lists?

A Because carry indicates the end of the linked list.

B Because carry resets the linked list pointers to start.

C Because carry might add an extra digit to the result linked list.

D Because carry stores the sum of all digits processed so far.

💡 Hint

Think about what happens if the last sum is 15.

If the last addition results in a carry (like 1), it means an extra digit must be added to the result linked list to represent that carry.

🔧 Debug

advanced

Identify the Bug in Linked List Addition Code

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

Code snippet:
while (l1 != NULL || l2 != NULL) { int sum = carry; if (l1 != NULL) { sum += l1->val; l1 = l1->next; } if (l2 != NULL) { sum += l2->val; l2 = l2->next; } carry = sum / 10; struct Node* newNode = malloc(sizeof(struct Node)); newNode->val = sum % 10; newNode->next = NULL; current->next = newNode; current = newNode; } // Missing carry check after loop

A Memory leak due to missing free calls.

B The last carry digit is lost, resulting in incorrect sum.

C Segmentation fault due to null pointer dereference.

D Infinite loop because loop condition never ends.

💡 Hint

What if the last sum is 15?

If carry is not zero after loop ends, it must be added as a new node. Missing this causes wrong result.

❓ Predict Output

advanced

Output of Adding Unequal Length Linked Lists

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

List 1: 9 -> 9 -> 9 -> 9 -> null (represents 9999)
List 2: 1 -> null (represents 1)

Use the standard addition code with carry.

A 9 -> 9 -> 9 -> 10 -> null

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

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

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

💡 Hint

Add digit by digit with carry, note length difference.

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

🚀 Application

expert

Calculate Length of Result Linked List After Addition

Given two linked lists representing numbers, what is the maximum possible length of the resulting linked list after addition?

A Maximum length is max(length of list1, length of list2) + 1

B Maximum length is sum of lengths of list1 and list2

C Maximum length is max(length of list1, length of list2)

D Maximum length is min(length of list1, length of list2) + 1

💡 Hint

Consider carry from the last digit addition.

Adding two numbers can produce a carry that adds one extra digit beyond the longer list length.