Insert at Specific Position in Doubly Linked List in DSA C++ - Time & Space Complexity

Time Complexity: Insert at Specific Position in Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to insert a node at a specific position in a doubly linked list changes as the list grows.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void insertAtPosition(Node*& head, int pos, int data) {
    Node* newNode = new Node(data);
    if (pos == 1) {
        newNode->next = head;
        if (head) head->prev = newNode;
        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;
    if (temp->next) temp->next->prev = newNode;
    temp->next = newNode;
    newNode->prev = temp;
}
    

This code inserts a new node at a given position in a doubly linked list by moving through the list to find the right spot.

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 gets longer, the number of steps to reach the insertion point grows roughly with the position number.

Input Size (n)	Approx. Operations
10	Up to 10 steps to reach position
100	Up to 100 steps to reach position
1000	Up to 1000 steps to reach position

Pattern observation: The time grows linearly with the position number, so doubling the list size roughly doubles the steps needed.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert depends on how far into the list we want to insert; more nodes mean more steps.

Common Mistake

[X] Wrong: "Inserting at any position in a doubly linked list always takes the same constant time."

[OK] Correct: You must first move to the insertion point, which takes time proportional to the position, so it is not always constant time.

Interview Connect

Understanding this helps you explain how linked lists work and why some operations are faster or slower, a key skill in many coding discussions.

Self-Check

"What if we kept a pointer to the tail of the list and wanted to insert near the end? How would the time complexity change?"

Practice

1. What happens when you insert a new node at position 1 in a doubly linked list?

easy

A. The list remains unchanged.

B. The new node becomes the new head of the list.

C. The new node replaces the last node.

D. The new node is added at the end of the list.

Solution

1. Step 1: Understand position 1 in doubly linked list

   Position 1 means the start of the list, so the new node should be the first node.

2. Step 2: Insert new node at head

   The new node's next pointer points to the old head, and the old head's previous pointer points back to the new node.

3. Final Answer:

   The new node becomes the new head of the list. -> Option B

4. Quick Check:

   Insert at position 1 = new head [OK]

Hint: Position 1 means insert at the start [OK]

Common Mistakes:

• Inserting at position 1 but not updating head pointer
• Not updating previous pointer of old head
• Confusing position 1 with end of list

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

easy

A. newNode->next = temp->next; temp->next->prev = newNode; temp->next = newNode; newNode->prev = temp;

B. newNode->prev = temp->prev; temp->prev->next = newNode; temp->prev = newNode; newNode->next = temp;

C. newNode->next = temp->prev; temp->prev->prev = newNode; temp->prev = newNode; newNode->prev = temp;

D. newNode->prev = temp->next; temp->next->next = newNode; temp->next = newNode; newNode->next = temp;

Solution

1. Step 1: Identify correct pointer updates for insertion

   When inserting after temp, newNode's next points to temp's next, and newNode's prev points to temp.

2. Step 2: Update surrounding nodes' pointers

   temp's next's prev must point to newNode, and temp's next must point to newNode.

3. Final Answer:

   newNode->next = temp->next; temp->next->prev = newNode; temp->next = newNode; newNode->prev = temp; -> Option A

4. Quick Check:

   Correct pointer updates = newNode->next = temp->next; temp->next->prev = newNode; temp->next = newNode; newNode->prev = temp; [OK]

Hint: Update next and prev pointers carefully [OK]

Common Mistakes:

• Mixing up next and prev pointers
• Not updating temp->next->prev pointer
• Assigning pointers in wrong order causing broken links

3. Given the doubly linked list: 10 <-> 20 <-> 30 <-> null, what will be the list after inserting 15 at position 2?

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

medium

A. 10 <-> 15 <-> 20 <-> 30 <-> null

B. 15 <-> 10 <-> 20 <-> 30 <-> null

C. 10 <-> 20 <-> 15 <-> 30 <-> null

D. 10 <-> 20 <-> 30 <-> 15 <-> null

Solution

1. Step 1: Locate position 2 in the list

   Position 2 means after the first node (10). So temp points to node with data 10.

2. Step 2: Insert new node with data 15 after temp

   New node's next points to 20, and prev points to 10. Update 10's next to new node and 20's prev to new node.

3. Final Answer:

   10 <-> 15 <-> 20 <-> 30 <-> null -> Option A

4. Quick Check:

   Insert at position 2 places new node after first node [OK]

Hint: Position 2 means insert after first node [OK]

Common Mistakes:

• Inserting before first node instead of after
• Not updating prev pointer of next node
• Skipping position count in loop

4. Identify the error in the following code snippet for inserting at position pos in a doubly linked list:

void insertAtPosition(Node*& head, int data, int pos) {
  Node* newNode = new Node(data);
  if (pos == 1) {
    newNode->next = head;
    if (head) head->prev = newNode;
    head = newNode;
    return;
  }
  Node* temp = head;
  for (int i = 1; i < pos - 1; i++) {
    temp = temp->next;
  }
  newNode->next = temp->next;
  temp->next = newNode;
  newNode->prev = temp;
  if (newNode->next) newNode->next->prev = newNode;
}

medium

A. The newNode->next pointer is assigned after temp->next is changed.

B. The newNode->prev pointer is not set correctly.

C. The head pointer is not updated when inserting at position 1.

D. The loop does not check if temp is nullptr, causing possible crash.

Solution

1. Step 1: Analyze the for loop traversal

   The loop moves temp forward but does not check if temp becomes nullptr, risking null pointer dereference.

2. Step 2: Identify missing safety check

   Without checking temp != nullptr, accessing temp->next can cause a crash if pos is invalid or list is short.

3. Final Answer:

   The loop does not check if temp is nullptr, causing possible crash. -> Option D

4. Quick Check:

   Always check temp != nullptr in traversal loops [OK]

Hint: Always check for nullptr when traversing [OK]

Common Mistakes:

• Ignoring nullptr checks in loops
• Not updating head for position 1 insert
• Incorrect order of pointer assignments

5. You have a doubly linked list: 5 <-> 10 <-> 15 <-> 20 <-> null. You want to insert a new node with value 12 at position 3. After insertion, what will be the correct sequence of the list? Also, explain how to handle insertion at the last position if the position is greater than the list length.

hard

A. 5 <-> 12 <-> 10 <-> 15 <-> 20 <-> null; Insert at last by replacing last node.

B. 5 <-> 10 <-> 15 <-> 12 <-> 20 <-> null; Insert at last by adding node before head.

C. 5 <-> 10 <-> 12 <-> 15 <-> 20 <-> null; Insert at last by traversing till last node and updating pointers.

D. 5 <-> 10 <-> 15 <-> 20 <-> 12 <-> null; Insert at last by ignoring position.

Solution

1. Step 1: Insert 12 at position 3

   Position 3 means after second node (10). Insert new node between 10 and 15, updating next and prev pointers accordingly.

2. Step 2: Handle insertion at last position if position > length

   Traverse to last node, then insert new node after it, updating last node's next and new node's prev pointers.

3. Final Answer:

   5 <-> 10 <-> 12 <-> 15 <-> 20 <-> null; Insert at last by traversing till last node and updating pointers. -> Option C

4. Quick Check:

   Insert at position 3 places node between 10 and 15; last insert by traversal [OK]

Hint: Insert at position or append if position too large [OK]

Common Mistakes:

• Inserting at wrong position
• Not updating prev pointer of next node
• Ignoring last position insertion logic