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

Practice - 5 Tasks

1fill in blank

easy

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

DSA C

struct Node* createNode(int val) {
    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = [1];
    newNode->next = NULL;
    return newNode;
}

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AnewNode

B0

CNULL

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning NULL or 0 instead of the passed value.

Assigning the pointer variable instead of the value.

We assign the passed value 'val' to the data field of the new node.

2fill in blank

medium

Complete the code to move to the next node in the linked list.

DSA C

while (current != NULL) {
    sum += current->data;
    current = [1];
}

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Acurrent->next

Bcurrent->prev

Ccurrent

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using current->prev which does not exist in singly linked list.

Assigning current to NULL prematurely.

To traverse the linked list, we move to the next node using current->next.

3fill in blank

hard

Fix the error in the code to correctly add carry to the sum digit.

DSA C

int sum = val1 + val2 + carry;
carry = [1];
sum = sum % 10;

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Asum / 10

Bcarry + 1

Csum % 10

Dval1 + val2

Attempts:

3 left

💡 Hint

Common Mistakes

Using modulo instead of division for carry.

Adding carry incorrectly.

Carry is the quotient when sum is divided by 10, so carry = sum / 10.

4fill in blank

hard

Fill both blanks to correctly create and link the new node in the result list.

DSA C

struct Node* newNode = createNode([1]);
if (resultHead == NULL) {
    resultHead = newNode;
    resultTail = newNode;
} else {
    resultTail->next = [2];
    resultTail = newNode;
}

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Asum

BnewNode

CresultHead

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Linking to resultHead instead of newNode.

Using NULL instead of newNode for next pointer.

We create a new node with the sum digit and link it by setting resultTail->next to newNode.

5fill in blank

hard

Fill all three blanks to correctly handle the loop and carry for adding two linked lists.

DSA C

while (l1 != NULL || l2 != NULL || [1]) {
    int val1 = (l1 != NULL) ? l1->data : 0;
    int val2 = (l2 != NULL) ? l2->data : 0;
    int sum = val1 + val2 + carry;
    carry = sum / 10;
    sum = sum % 10;
    struct Node* newNode = createNode([2]);
    if (resultHead == NULL) {
        resultHead = newNode;
        resultTail = newNode;
    } else {
        resultTail->next = newNode;
        resultTail = newNode;
    }
    if (l1 != NULL) l1 = l1->next;
    if (l2 != NULL) l2 = l2->next;
}
return [3];

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Acarry

Bsum

CresultHead

DresultTail

Attempts:

3 left

💡 Hint

Common Mistakes

Not checking carry in loop condition.

Returning tail instead of head.

Creating node with carry instead of sum.

Loop continues while carry exists; new node holds sum digit; return head of result list.