Creating a Singly Linked List from Scratch in DSA C - Code Step by Step

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the node structure for a singly linked list.

DSA C

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

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Apointer

Bprev

Clink

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

Using generic names like 'pointer' or 'link' which are uncommon.

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

2fill in blank

medium

Complete the code to create a new node dynamically.

DSA C

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

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ANode

Bnode

Cstruct Node

DNode*

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'node' which is undefined.

Using 'struct Node' inside sizeof without typedef.

We allocate memory for the Node structure using sizeof(Node).

3fill in blank

hard

Fix the error in the function to append a node at the end of the list.

DSA C

void appendNode(Node** head, int value) {
    Node* newNode = createNode(value);
    if (*head == NULL) {
        *head = newNode;
        return;
    }
    Node* temp = *head;
    while (temp->[1] != NULL) {
        temp = temp->next;
    }
    temp->next = newNode;
}

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Anext

Bprev

Clink

Dpointer

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

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

To traverse the list, we follow the next pointer until it is NULL.

4fill in blank

hard

Fill both blanks to insert a node at the beginning of the list.

DSA C

void insertAtHead(Node** head, int value) {
    Node* newNode = createNode(value);
    newNode->[1] = *head;
    *head = [2];
}

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Anext

Bhead

CnewNode

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is not used in singly linked lists.

Assigning head to newNode->next instead of the other way.

The new node's next points to the current head, then head is updated to the new node.

5fill in blank

hard

Fill all three blanks to print all elements of the singly linked list.

DSA C

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

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ANULL

Bdata

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' instead of NULL in the loop condition.

Printing 'next' instead of 'data'.

Moving current to 'data' instead of 'next'.

We loop until current is NULL, print current's data, then move to current->next.

Practice

1. What is the main purpose of a singly linked list in data structures?

easy

A. To store elements with direct access by index like an array

B. To store elements in a sequence where each element points to the next

C. To store elements in a sorted order automatically

D. To store elements with multiple pointers to previous nodes

Solution

1. Step 1: Understand singly linked list structure

   A singly linked list stores elements where each node points only to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike arrays, singly linked lists do not provide direct index access, and unlike doubly linked lists, nodes have only one pointer.

3. Final Answer:

   To store elements in a sequence where each element points to the next -> Option B

4. Quick Check:

   Singly linked list = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes have one next pointer only [OK]

Common Mistakes:

• Thinking singly linked list allows direct index access
• Confusing singly linked list with doubly linked list
• Assuming automatic sorting in singly linked list

2. Which of the following is the correct way to define a node structure for a singly linked list in C?

easy

A. typedef struct Node { int data; struct Node* next; } Node;

B. typedef struct Node { int data; Node* next; } Node;

C. struct Node { int data; struct Node next; };

D. struct Node { int data; struct Node* prev; };

Solution

1. Step 1: Check pointer usage in struct

   The next pointer must be a pointer to struct Node, so use struct Node* next;.

2. Step 2: Verify typedef correctness

   typedef struct Node { int data; struct Node* next; } Node; correctly uses typedef and pointer syntax for next node.

3. Final Answer:

   typedef struct Node { int data; struct Node* next; } Node; -> Option A

4. Quick Check:

   Node struct uses pointer to struct Node for next [OK]

Hint: Use 'struct Node* next;' inside struct for next pointer [OK]

Common Mistakes:

• Omitting 'struct' keyword before Node* inside struct
• Using 'Node* next;' without typedef defined yet
• Using 'struct Node next;' instead of pointer
• Using 'prev' pointer instead of 'next' for singly linked list

3. What will be the output of the following C code snippet that creates a singly linked list with two nodes and prints their data?

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* head = malloc(sizeof(Node));
    head->data = 10;
    head->next = malloc(sizeof(Node));
    head->next->data = 20;
    head->next->next = NULL;

    Node* temp = head;
    while (temp != NULL) {
        printf("%d ", temp->data);
        temp = temp->next;
    }
    return 0;
}

medium

A. 10 20

B. 20 10

C. 10

D. Compilation error

Solution

1. Step 1: Trace node creation and data assignment

   First node data is 10, second node data is 20, linked correctly.

2. Step 2: Trace printing loop

   Loop prints data of head (10), then moves to next node (20), then stops at NULL.

3. Final Answer:

   10 20 -> Option A

4. Quick Check:

   Prints nodes in order: 10 then 20 [OK]

Hint: Follow next pointers to print nodes in order [OK]

Common Mistakes:

• Confusing order of nodes printed
• Forgetting to set next of last node to NULL
• Assuming compilation error due to missing includes (not shown)

4. Identify the error in the following code that attempts to create a singly linked list node and assign data:

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* head;
    head->data = 5;
    head->next = NULL;
    return 0;
}

medium

A. data should be a pointer, not int

B. next pointer should point to previous node

C. head is not initialized before accessing its members

D. struct Node definition is incorrect

Solution

1. Step 1: Check pointer initialization

   head is declared but not assigned memory before accessing its members.

2. Step 2: Understand consequences

   Accessing head->data without allocation causes undefined behavior or crash.

3. Final Answer:

   head is not initialized before accessing its members -> Option C

4. Quick Check:

   Pointer must be allocated before use [OK]

Hint: Always allocate memory before using pointer members [OK]

Common Mistakes:

• Assuming pointer is auto-initialized
• Confusing next pointer direction
• Thinking data must be pointer type

5. You want to create a singly linked list from scratch that stores integers 1, 2, 3 in order. Which sequence of operations correctly builds the list?

hard

A. Allocate node1, set data=1, set node1->next=NULL, allocate node2, set data=2, set node2->next=NULL, allocate node3, set data=3, set node3->next=NULL

B. Allocate node1, set data=1, allocate node2, set data=2, allocate node3, set data=3, link node3->next=node2, link node2->next=node1, set node1->next=NULL

C. Allocate node1, set data=1, link node1->next=node2, allocate node2, set data=2, link node2->next=node3, allocate node3, set data=3, set node3->next=NULL

D. Allocate node1, set data=1, allocate node2, set data=2, link node1->next=node2, allocate node3, set data=3, link node2->next=node3, set node3->next=NULL

Solution

1. Step 1: Allocate nodes and assign data in order

   Nodes must be allocated and data assigned before linking.

2. Step 2: Link nodes in correct sequence

   Link node1 to node2, node2 to node3, and node3 to NULL to end list.

3. Final Answer:

   Allocate node1, set data=1, allocate node2, set data=2, link node1->next=node2, allocate node3, set data=3, link node2->next=node3, set node3->next=NULL -> Option D

4. Quick Check:

   Correct order and linking builds list 1->2->3->NULL [OK]

Hint: Link nodes only after allocating and setting data [OK]

Common Mistakes:

• Linking nodes before allocation
• Not setting last node's next to NULL
• Linking nodes in reverse order