Why Linked List Exists and What Problem It Solves in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to the next node in a linked list node structure.

DSA C

struct Node {
    int data;
    struct Node* [1];
};

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Anext

Bprev

Clink

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using generic names like 'pointer' which are unclear.

The pointer to the next node in a linked list is commonly named next.

2fill in blank

medium

Complete the code to create a new node dynamically in a linked list.

DSA C

struct Node* newNode = (struct Node*) malloc(sizeof([1]));

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Astruct

BNode

Cnode

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'node' which is undefined.

Using 'int' which is the wrong type.

We allocate memory for a Node structure, so sizeof(Node) is correct.

3fill in blank

hard

Fix the error in the code to properly link a new node to the existing linked list.

DSA C

newNode->[1] = head;

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Apointer

Bprev

Clink

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using undefined pointer names like 'link' or 'pointer'.

The new node's next pointer should point to the current head to insert at the front.

4fill in blank

hard

Fill both blanks to complete the function that inserts a new node at the beginning of the linked list.

DSA C

void insertAtFront(struct Node** head_ref, int new_data) {
    struct Node* newNode = (struct Node*) malloc(sizeof([1]));
    newNode->data = new_data;
    newNode->[2] = *head_ref;
    *head_ref = newNode;
}

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ANode

Bnode

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'node' which is undefined.

Using 'prev' which is for doubly linked lists.

Memory is allocated for a Node, and the new node's next pointer is set to the current head.

5fill in blank

hard

Fill all three blanks to complete the function that prints all elements of a linked list.

DSA C

void printList(struct Node* [1]) {
    struct Node* [2] = [3];
    while (temp != NULL) {
        printf("%d -> ", temp->data);
        temp = temp->next;
    }
    printf("NULL\n");
}

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Ahead

Btemp

Ccurrent

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same name for parameter and temp pointer causing confusion.

Using undefined variable names in the loop.

The function takes a pointer named head. A temporary pointer temp is set to head to traverse the list.