Insert at Beginning Head Insert in DSA C++ - Time & Space Complexity

Time Complexity: Insert at Beginning Head Insert

O(1)

Understanding Time Complexity

We want to understand how long it takes to add a new item at the start of a linked list.

How does the time needed change as the list grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

void insertAtBeginning(Node*& head, int value) {
    Node* newNode = new Node();
    newNode->data = value;
    newNode->next = head;
    head = newNode;
}
    

This code adds a new node at the start of a linked list by adjusting pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Creating and linking one new node.
• How many times: Exactly once per insertion, no loops or recursion.

How Execution Grows With Input

Adding a node at the start takes the same steps no matter how big the list is.

Input Size (n)	Approx. Operations
10	4 steps (create node, assign data, link, update head)
100	4 steps (same as above)
1000	4 steps (same as above)

Pattern observation: The number of steps stays the same regardless of list size.

Final Time Complexity

Time Complexity: O(1)

This means adding at the start always takes the same short time, no matter how big the list is.

Common Mistake

[X] Wrong: "Adding at the start takes longer as the list grows because the list is bigger."

[OK] Correct: We only change a few pointers at the start, so the list size does not affect the time.

Interview Connect

Knowing that head insertion is quick helps you choose the right method when speed matters.

Self-Check

"What if we had to insert at the end of the list instead? How would the time complexity change?"

Practice

1. What does the "Insert at Beginning" operation do in a singly linked list?

easy

A. Adds a new node at the end of the list

B. Deletes the first node of the list

C. Adds a new node at the start of the list, making it the new head

D. Searches for a node at the beginning

Solution

1. Step 1: Understand the operation

   Inserting at the beginning means placing a new node before the current head.

2. Step 2: Effect on the list

   The new node becomes the head, and its next pointer points to the old head.

3. Final Answer:

   Adds a new node at the start of the list, making it the new head -> Option C

4. Quick Check:

   Insert at beginning = new head [OK]

Hint: New node always becomes the head in beginning insert [OK]

Common Mistakes:

• Confusing insert at beginning with insert at end
• Thinking it deletes nodes
• Assuming it searches instead of inserts

2. Which of the following is the correct syntax to insert a new node at the beginning of a singly linked list in C++?

easy

A. head->next = newNode; newNode = head;

B. newNode->next = head; head = newNode;

C. head = newNode; newNode->next = head;

D. newNode = head; head = newNode->next;

Solution

1. Step 1: Link new node to current head

   Set newNode->next to point to the current head node.

2. Step 2: Update head to new node

   Assign head to newNode so it becomes the new first node.

3. Final Answer:

   newNode->next = head; head = newNode; -> Option B

4. Quick Check:

   Link then update head [OK]

Hint: Set newNode->next first, then update head [OK]

Common Mistakes:

• Updating head before linking newNode->next
• Swapping pointers incorrectly
• Assigning newNode to head->next

3. What will be the output linked list after executing this code?

struct Node { int data; Node* next; };
Node* head = nullptr;
Node* newNode = new Node{10, nullptr};
newNode->next = head;
head = newNode;
newNode = new Node{20, nullptr};
newNode->next = head;
head = newNode;

Print the list from head to end.

medium

A. 20 -> 10 -> null

B. 10 -> 20 -> null

C. null -> 10 -> 20

D. 10 -> null

Solution

1. Step 1: Insert first node with data 10

   head points to node with 10, next is nullptr.

2. Step 2: Insert second node with data 20 at beginning

   newNode->next = head (node 10), head = newNode (node 20).

3. Final Answer:

   20 -> 10 -> null -> Option A

4. Quick Check:

   Last inserted node is head [OK]

Hint: Last inserted node is always head [OK]

Common Mistakes:

• Assuming order is insertion order instead of reversed
• Confusing next pointers
• Ignoring nullptr termination

4. Identify the error in this code snippet for inserting at the beginning of a linked list:

Node* newNode = new Node();
newNode->data = 5;
head = newNode;
newNode->next = head;

medium

A. No error, code is correct

B. newNode->data is not initialized

C. head should be set to nullptr first

D. newNode->next should be assigned before head is updated

Solution

1. Step 1: Check pointer assignments order

   Assigning head = newNode before setting newNode->next causes newNode->next to point to itself.

2. Step 2: Correct order

   newNode->next should be set to current head before updating head to newNode.

3. Final Answer:

   newNode->next should be assigned before head is updated -> Option D

4. Quick Check:

   Link newNode->next before head update [OK]

Hint: Always link newNode->next before changing head [OK]

Common Mistakes:

• Updating head before linking newNode->next
• Forgetting to initialize newNode->data
• Assuming no error in pointer assignments

5. Given a linked list with nodes 30 -> 20 -> 10 -> null, what will be the list after inserting a new node with data 40 at the beginning twice in a row?

hard

A. 40 -> 40 -> 30 -> 20 -> 10 -> null

B. 30 -> 20 -> 10 -> 40 -> 40 -> null

C. 40 -> 30 -> 20 -> 10 -> null

D. 40 -> 30 -> 20 -> 10 -> 40 -> null

Solution

1. Step 1: Insert first 40 at beginning

   New head is 40, next points to old head 30.

2. Step 2: Insert second 40 at beginning

   New head is another 40, next points to previous 40 node.

3. Final Answer:

   40 -> 40 -> 30 -> 20 -> 10 -> null -> Option A

4. Quick Check:

   Each insert adds new head [OK]

Hint: Each insert adds a new head node [OK]

Common Mistakes:

• Assuming duplicate nodes merge
• Inserting at end instead of beginning
• Losing track of next pointers