Insert at Middle Specific Position in DSA C++ - Time & Space Complexity

Time Complexity: Insert at Middle Specific Position

O(n)

Understanding Time Complexity

When inserting an element at a specific middle position in a linked list, we want to know how the time needed changes as the list grows.

We ask: How does the work increase when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void insertAtPosition(Node*& head, int data, int pos) {
    Node* newNode = new Node(data);
    if (pos == 1) {
        newNode->next = head;
        head = newNode;
        return;
    }
    Node* temp = head;
    for (int i = 1; i < pos - 1 && temp != nullptr; i++) {
        temp = temp->next;
    }
    if (temp == nullptr) return; // position out of bounds
    newNode->next = temp->next;
    temp->next = newNode;
}
    

This code inserts a new node at a given position in a singly linked list by first moving to the node just before the position.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop that moves through the list nodes to reach the position before insertion.
• How many times: Up to pos - 1 times, which depends on the position where we insert.

How Execution Grows With Input

As the list grows, the time to reach the middle position grows roughly in proportion to the position number.

Input Size (n)	Approx. Operations
10	About 5 steps to reach middle
100	About 50 steps to reach middle
1000	About 500 steps to reach middle

Pattern observation: The steps increase linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert grows in a straight line with the size of the list.

Common Mistake

[X] Wrong: "Inserting at a middle position is always fast and takes constant time."

[OK] Correct: Because we must first move through the list to find the position, which takes time proportional to the position number.

Interview Connect

Understanding how insertion time grows helps you explain linked list behavior clearly and shows you can reason about code efficiency in real situations.

Self-Check

"What if we changed the linked list to a doubly linked list? How would the time complexity for insertion at a middle position change?"

Practice

1. What does inserting a node at a specific middle position in a linked list mean?

easy

A. Adding a new node only at the start of the list

B. Adding a new node exactly at the given position counting from the head

C. Adding a new node only at the end of the list

D. Replacing the node at the given position with a new node

Solution

1. Step 1: Understand insertion position

   Inserting at a specific middle position means placing the new node exactly at that position counting from the start (head) of the list.

2. Step 2: Differentiate from other insertions

   Insertion at start or end is different; here we insert in between nodes without replacing any node.

3. Final Answer:

   Adding a new node exactly at the given position counting from the head -> Option B

4. Quick Check:

   Insert at position means place new node there [OK]

Hint: Insert means add new node at exact position, not replace [OK]

Common Mistakes:

• Confusing insertion with replacement
• Thinking insertion only happens at start or end
• Ignoring position counting from head

2. Which of the following C++ code snippets correctly inserts a new node with value 10 at position 3 in a singly linked list?

easy

A. Node* temp = head; for(int i=1; i<2; i++) temp = temp->next; Node* newNode = new Node(10); newNode->next = temp->next; temp->next = newNode;

B. Node* newNode = new Node(10); newNode->next = head; head = newNode;

C. Node* temp = head; for(int i=1; i<3; i++) temp = temp->next; Node* newNode = new Node(10); newNode->next = temp->next; temp->next = newNode;

D. Node* newNode = new Node(10); head->next = newNode; newNode->next = NULL;

Solution

1. Step 1: Identify correct loop for position 3 insertion

   To insert at position 3, we stop at node 2 (position 2), so loop runs i=1 to i<2.

2. Step 2: Check pointer updates

   New node's next points to temp->next, then temp->next updated to new node. This correctly inserts at position 3.

3. Final Answer:

   Code snippet in option A correctly inserts at position 3 -> Option A

4. Quick Check:

   Loop stops at position before insertion [OK]

Hint: Loop till position-1, then insert new node [OK]

Common Mistakes:

• Looping one extra time causing wrong position
• Not updating newNode->next before temp->next
• Inserting at wrong position due to loop error

3. What will be the linked list after inserting value 15 at position 2 in the list: 5 -> 10 -> 20 -> null?

Node* head = new Node(5);
head->next = new Node(10);
head->next->next = new Node(20);

// Insert 15 at position 2
Node* temp = head;
for(int i=1; i<1; i++) temp = temp->next;
Node* newNode = new Node(15);
newNode->next = temp->next;
temp->next = newNode;

medium

A. 5 -> 15 -> 10 -> 20 -> null

B. 15 -> 5 -> 10 -> 20 -> null

C. 5 -> 10 -> 15 -> 20 -> null

D. 5 -> 10 -> 20 -> 15 -> null

Solution

1. Step 1: Trace loop to find insertion point

   Loop runs i=1 to i<1, so no iteration. temp points to head (value 5).

2. Step 2: Insert new node after temp

   New node 15 points to temp->next (10), then temp->next points to new node. List becomes 5 -> 15 -> 10 -> 20 -> null.

3. Final Answer:

   5 -> 15 -> 10 -> 20 -> null -> Option A

4. Quick Check:

   Insert after node 1 gives new node at position 2 [OK]

Hint: Loop zero times means insert after head [OK]

Common Mistakes:

• Miscounting loop iterations
• Inserting at wrong position
• Not updating pointers correctly

4. Identify the error in the following code that tries to insert a node with value 25 at position 4 in a linked list:

Node* temp = head;
for(int i=1; i<=4; i++) {
  temp = temp->next;
}
Node* newNode = new Node(25);
newNode->next = temp->next;
temp->next = newNode;

medium

A. No error, code works fine

B. New node's next pointer is not set correctly

C. New node is not created properly

D. Loop runs too many times causing temp to become NULL

Solution

1. Step 1: Analyze loop iterations

   Loop runs i=1 to i<=4 (4 times), moving temp 4 nodes ahead. For insertion at position 4, temp should stop at position 3.

2. Step 2: Consequence of loop error

   Temp moves too far, possibly NULL, causing runtime error when accessing temp->next.

3. Final Answer:

   Loop runs too many times causing temp to become NULL -> Option D

4. Quick Check:

   Loop must run position-2 times, not position times [OK]

Hint: Loop till position-2 to avoid NULL pointer [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Not checking for NULL before accessing next
• Incorrect loop range for insertion

5. Given a singly linked list with nodes: 1 -> 3 -> 5 -> 7 -> null, write the steps to insert a new node with value 4 at position 3. What will be the final list?

hard

A. 1 -> 4 -> 3 -> 5 -> 7 -> null

B. 4 -> 1 -> 3 -> 5 -> 7 -> null

C. 1 -> 3 -> 4 -> 5 -> 7 -> null

D. 1 -> 3 -> 5 -> 7 -> 4 -> null

Solution

1. Step 1: Traverse to node before position 3

   Start at head (1), move to next node (3) which is position 2. Stop here to insert after this node.

2. Step 2: Insert new node with value 4

   Create new node 4, set its next to current node's next (5), then update current node's next to new node.

3. Final Answer:

   1 -> 3 -> 4 -> 5 -> 7 -> null -> Option C

4. Quick Check:

   Insert at position 3 places new node between 3 and 5 [OK]

Hint: Stop at position-1 node, insert new node after it [OK]

Common Mistakes:

• Inserting at wrong position
• Not updating pointers correctly
• Confusing position counting from 0 or 1