Creating a Singly Linked List from Scratch in DSA C - Complexity Walkthrough

Time Complexity: Creating a Singly Linked List from Scratch

O(n)

Understanding Time Complexity

When we create a singly linked list from scratch, we add nodes one by one. Understanding time complexity helps us see how the work grows as we add more nodes.

We want to know: How much time does it take to build the whole list as it gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

struct Node* createList(int arr[], int n) {
    struct Node* head = NULL;
    struct Node* tail = NULL;
    for (int i = 0; i < n; i++) {
        struct Node* newNode = malloc(sizeof(struct Node));
        newNode->data = arr[i];
        newNode->next = NULL;
        if (head == NULL) {
            head = newNode;
            tail = newNode;
        } else {
            tail->next = newNode;
            tail = newNode;
        }
    }
    return head;
}
    

This code creates a singly linked list by adding each element from an array to the end of the list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs through each element of the input array once.
• How many times: Exactly n times, where n is the number of elements.

How Execution Grows With Input

Each new node is created and linked once per element, so the work grows directly with the number of elements.

Input Size (n)	Approx. Operations
10	About 10 node creations and links
100	About 100 node creations and links
1000	About 1000 node creations and links

Pattern observation: The number of operations grows linearly as the input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to create the list grows in direct proportion to the number of elements we add.

Common Mistake

[X] Wrong: "Adding each node to the end takes constant time, so the whole list creation is faster than linear."

[OK] Correct: While adding to the end is constant time here because we keep a tail pointer, if we did not keep track of the tail, each addition would require traversing the list, making it slower.

Interview Connect

Knowing how list creation scales helps you explain your code clearly and shows you understand how data structures work under the hood. This skill is useful in many coding tasks and interviews.

Self-Check

"What if we did not keep a tail pointer and instead found the end of the list each time we added a node? How would the time complexity change?"

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