LINK01P04 - Editorial

editorial
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Problem Link - Optimal insertion at the end

Problem Statement:

By maintaining a tail pointer, which will point to the last element of the linked list. Thus whenever we want to insert at the end, we can use tail for that.

Task
Add a new tail pointer and update the current insertAtEnd to pass this exercise.

Approach:

The key idea of this solution is to efficiently insert elements at the end of the list using a tail pointer. By maintaining both the head and tail pointers, we can avoid traversing the entire list each time we want to add a new node, making the insertion process faster.

Here’s how the approach works:

  1. Node Structure:

    • A Node stores two things:
      • value: The data of the node.
      • next: A pointer that points to the next node in the list (or nullptr if it’s the last node).

  2. LinkedList Class:

    • head: A pointer to the first node in the list.

    • tail: A pointer to the last node in the list.

  3. insertAtEnd(int value):

    • This function adds a new node to the end of the list.

    • If the list is empty (head == nullptr), the new node becomes both the head and tail.

    • Otherwise, the new node is linked to the tail, and then the tail is updated to point to this new node.

  4. printValues():

    • This function prints all the values of the nodes in the list from the head to the tail.

The code uses this approach to efficiently manage the linked list.

Time Complexity:

  • insertAtEnd: O(1) since we directly update the tail pointer, avoiding traversal of the list.

  • printValues: O(n) where n is the number of nodes, as we need to traverse the entire list to print all elements.

Space Complexity:

  • O(n) since we store n nodes in the linked list, with each node storing one value and a pointer to the next node.