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

When inserting a new node at a specific position in a doubly linked list, we want to know how the time needed changes as the list grows.

We ask: How many steps does it take to find the right spot and insert the node?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void insertAtPosition(Node** head, int data, int pos) {
    Node* newNode = malloc(sizeof(Node));
    newNode->data = data;
    newNode->prev = NULL;
    newNode->next = NULL;

    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 != NULL; i++) {
        temp = temp->next;
    }

    if (temp == NULL) 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 by first moving through the list to find the correct spot, then adjusting pointers.

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 - 2) times, depending on the position where we insert.

How Execution Grows With Input

As the list gets longer, the time to reach the insertion point grows roughly in direct proportion to the position number.

Input Size (n)	Approx. Operations
10	Up to 9 steps to reach position
100	Up to 99 steps to reach position
1000	Up to 999 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 grows in a straight line with the list size or the position where we insert.

Common Mistake

[X] Wrong: "Inserting at any position in a doubly linked list always takes constant time O(1)."

[OK] Correct: You must first find the position by moving through the list, which takes time proportional to the position number, so it is not always instant.

Interview Connect

Understanding how insertion time depends on position helps you explain trade-offs in linked list operations clearly and confidently.

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 new node becomes the new head of the list.

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

C. The new node replaces the last node.

D. The list remains unchanged.

Solution

1. Step 1: Understand position 1 insertion

   Inserting at position 1 means adding the node at the start of the list.

2. Step 2: Effect on head pointer

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

3. Final Answer:

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

4. Quick Check:

   Position 1 insertion updates head [OK]

Hint: Position 1 means new head insertion [OK]

Common Mistakes:

• Thinking position 1 means end insertion
• Not updating the head pointer
• Ignoring previous pointer updates

2. Which of the following is the correct way to update pointers when inserting a node at position 3 in a doubly linked list?

easy

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

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

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

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

Solution

1. Step 1: Identify insertion point

   At position 3, newNode is inserted after temp (node at position 2).

2. Step 2: Update pointers correctly

   newNode's next points to temp's next, prev to temp; temp's next's prev updated to newNode; temp's next updated to newNode.

3. Final Answer:

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

4. Quick Check:

   Correct pointer updates for middle insertion [OK]

Hint: Update newNode next/prev before changing temp pointers [OK]

Common Mistakes:

• Swapping next and prev pointers
• Not updating temp->next->prev
• Forgetting to update temp->next

3. What will be the list after inserting a node with value 10 at position 2 in the doubly linked list: 5 <-> 15 <-> 20?

medium

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

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

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

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

Solution

1. Step 1: Locate position 2

   Position 2 means new node goes between nodes 1 (5) and 2 (15).

2. Step 2: Insert new node with value 10

   New node's prev points to 5, next points to 15; update 5's next and 15's prev accordingly.

3. Final Answer:

   5 <-> 10 <-> 15 <-> 20 -> Option D

4. Quick Check:

   Inserted node at position 2 between 5 and 15 [OK]

Hint: Position 2 means between first and second nodes [OK]

Common Mistakes:

• Inserting at wrong position
• Not updating both next and prev pointers
• Appending instead of inserting

4. Given the code snippet below, what is the error when inserting at position 4 in a list of 3 nodes?

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

medium

A. newNode is not allocated properly.

B. temp->next is NULL, so temp->next->prev causes a runtime error.

C. Loop runs one time too many causing segmentation fault.

D. prev pointer of newNode is not set.

Solution

1. Step 1: Analyze loop for position 4 in 3-node list

   Loop moves temp to node at position 3; temp->next is NULL (end of list).

2. Step 2: Check pointer updates

   Accessing temp->next->prev causes runtime error because temp->next is NULL.

3. Final Answer:

   temp->next is NULL, so temp->next->prev causes a runtime error. -> Option B

4. Quick Check:

   Accessing NULL pointer causes runtime error [OK]

Hint: Check for NULL before accessing next->prev [OK]

Common Mistakes:

• Not checking if temp->next is NULL
• Assuming list is longer than it is
• Ignoring edge case at list end

5. You have a doubly linked list: 1 <-> 3 <-> 5 <-> 7. You want to insert a node with value 4 at position 3. After insertion, what will be the correct list and how do you handle the pointers to maintain list integrity?

hard

A. 1 <-> 3 <-> 4 <-> 5 <-> 7; Set newNode->prev = node at pos 2, newNode->next = node at pos 3, update adjacent nodes' pointers.

B. 1 <-> 3 <-> 5 <-> 4 <-> 7; Insert newNode after node at pos 3 without updating prev pointers.

C. 1 <-> 4 <-> 3 <-> 5 <-> 7; Insert newNode at head and shift others.

D. 1 <-> 3 <-> 5 <-> 7 <-> 4; Append newNode at end regardless of position.

Solution

1. Step 1: Identify insertion position and nodes

   Position 3 means new node goes between nodes with values 3 (pos 2) and 5 (pos 3).

2. Step 2: Update pointers to insert new node

   Set newNode->prev to node with value 3, newNode->next to node with value 5; update node 3's next and node 5's prev to newNode.

3. Step 3: Verify list integrity

   After insertion, list is 1 <-> 3 <-> 4 <-> 5 <-> 7 with all pointers correctly linked.

4. Final Answer:

   1 <-> 3 <-> 4 <-> 5 <-> 7; Set newNode->prev = node at pos 2, newNode->next = node at pos 3, update adjacent nodes' pointers. -> Option A

5. Quick Check:

   Proper pointer updates maintain doubly linked list [OK]

Hint: Insert between nodes, update both next and prev pointers [OK]

Common Mistakes:

• Inserting at wrong position
• Not updating both adjacent nodes' pointers
• Breaking list links causing traversal errors