Insert at End of Doubly Linked List in DSA C++ - Time & Space Complexity

Time Complexity: Insert at End of Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time to insert a new node at the end of a doubly linked list changes as the list grows.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

void insertAtEnd(Node*& head, int val) {
    Node* newNode = new Node{val, nullptr, nullptr};
    if (!head) {
        head = newNode;
        return;
    }
    Node* temp = head;
    while (temp->next) {
        temp = temp->next;
    }
    temp->next = newNode;
    newNode->prev = temp;
}
    

This code inserts a new node with value val at the end of a doubly linked list starting at head.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through the list to find the last node.
• How many times: It runs once for each node until the end, so roughly n times if the list has n nodes.

How Execution Grows With Input

As the list gets longer, the loop takes more steps to reach the end.

Input Size (n)	Approx. Operations
10	About 10 steps to reach the end
100	About 100 steps to reach the end
1000	About 1000 steps to reach the end

Pattern observation: The number of steps grows directly with the size of the list.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the end grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Inserting at the end is always fast and takes constant time."

[OK] Correct: Without a pointer to the last node, you must walk through the whole list, which takes longer as the list grows.

Interview Connect

Understanding this helps you explain how linked lists work and why keeping track of the tail node can speed up insertions.

Self-Check

"What if we kept a pointer to the last node (tail) of the list? How would the time complexity change?"

Practice

1. What is the main purpose of inserting a node at the end of a doubly linked list?

easy

A. To add a new element after the last node

B. To remove the first node

C. To reverse the list

D. To find the middle node

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes connected forward and backward, allowing traversal in both directions.

2. Step 2: Purpose of insertion at the end

   Inserting at the end means adding a new node after the current last node, updating pointers accordingly.

3. Final Answer:

   To add a new element after the last node -> Option A

4. Quick Check:

   Insert at end = Add after last node [OK]

Hint: Insert at end means add after last node [OK]

Common Mistakes:

• Confusing insertion at end with deletion
• Thinking insertion at end reverses the list
• Assuming insertion happens at the start

2. Which of the following is the correct way to update the tail pointer when inserting a new node at the end of a doubly linked list in C++?

easy

A. tail = NULL;

B. tail = tail->prev;

C. tail = head;

D. tail = newNode;

Solution

1. Step 1: Identify the new tail after insertion

   When a new node is added at the end, it becomes the new tail of the list.

2. Step 2: Update the tail pointer

   Assign the tail pointer to point to the new node to reflect the updated end of the list.

3. Final Answer:

   tail = newNode; -> Option D

4. Quick Check:

   New tail points to newNode [OK]

Hint: Tail always points to the last inserted node [OK]

Common Mistakes:

• Setting tail to previous node instead of new node
• Resetting tail to head incorrectly
• Assigning tail to NULL after insertion

3. What will be the output of the following code snippet after inserting 30 at the end of the doubly linked list?

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

Node* head = new Node{10, NULL, NULL};
Node* tail = head;

Node* newNode = new Node{30, tail, NULL};
tail->next = newNode;
tail = newNode;

// Print list from head
Node* temp = head;
while(temp != NULL) {
  std::cout << temp->data << " ";
  temp = temp->next;
}

medium

A. 10 30

B. 30 10

C. 10

D. 30

Solution

1. Step 1: Initial list and insertion

   Initially, list has one node with data 10. New node with 30 is added at end, linked properly.

2. Step 2: Traversing and printing

   Starting from head (10), print data then move to next node (30), print data, then NULL ends loop.

3. Final Answer:

   10 30 -> Option A

4. Quick Check:

   List after insertion = 10 30 [OK]

Hint: Traverse from head to see all nodes in order [OK]

Common Mistakes:

• Printing nodes in reverse order
• Not updating tail->next pointer
• Forgetting to update tail pointer

4. Identify the error in the following code snippet that inserts a node at the end of a doubly linked list:

void insertEnd(Node*& head, Node*& tail, int val) {
  Node* newNode = new Node{val, NULL, NULL};
  if (head == NULL) {
    head = newNode;
    tail = newNode;
  } else {
    tail->next = newNode;
    newNode->prev = tail;
    head = newNode;  // Error here
  }
}

medium

A. head should be updated to newNode

B. newNode->prev should be NULL

C. tail should be updated to newNode

D. tail->next should be NULL

Solution

1. Step 1: Analyze pointer updates in else block

   tail->next points to newNode and newNode->prev points to tail correctly.

2. Step 2: Identify incorrect pointer assignment

   head is wrongly updated to newNode; it should remain unchanged. Instead, tail should be updated to newNode.

3. Final Answer:

   tail should be updated to newNode -> Option C

4. Quick Check:

   Tail points to last node after insertion [OK]

Hint: Update tail, not head, when inserting at end [OK]

Common Mistakes:

• Updating head instead of tail
• Not linking newNode's prev pointer
• Forgetting to set tail->next

5. Given a doubly linked list with nodes 5 <-> 10 <-> 15, write the sequence of node values after inserting 20 at the end and then 2 at the end. What is the final list?

hard

A. 2 <-> 5 <-> 10 <-> 15 <-> 20

B. 5 <-> 10 <-> 15 <-> 20 <-> 2

C. 5 <-> 10 <-> 15 <-> 2 <-> 20

D. 20 <-> 2 <-> 5 <-> 10 <-> 15

Solution

1. Step 1: Insert 20 at the end

   Starting list: 5 <-> 10 <-> 15. After inserting 20 at end: 5 <-> 10 <-> 15 <-> 20.

2. Step 2: Insert 2 at the end

   Now add 2 at end: 5 <-> 10 <-> 15 <-> 20 <-> 2.

3. Final Answer:

   5 <-> 10 <-> 15 <-> 20 <-> 2 -> Option B

4. Quick Check:

   Insertions add nodes at list end in order [OK]

Hint: Insertions add nodes at the end, order preserved [OK]

Common Mistakes:

• Inserting new nodes at the start instead of end
• Mixing order of inserted nodes
• Confusing next and prev pointers