What if you could remove the last item instantly without searching the whole list?
Why Delete from End of Doubly Linked List in DSA C#?
Imagine you have a long chain of paper clips linked together. You want to remove the last paper clip, but you only have access to the first one. You try to find the last clip by checking each one one-by-one from the start.
This manual way is slow because you must go through every paper clip until you reach the end. It is easy to lose track or make mistakes, especially if the chain is very long. Removing the last clip without a clear way to jump to it wastes time and effort.
A doubly linked list keeps track of both the start and the end of the chain. This lets you quickly find and remove the last item without checking every single one. The links go both ways, so you can easily update the chain after removing the last clip.
Node current = head;
while (current.next != null) {
current = current.next;
}
// Remove current nodeif (tail != null) { Node toRemove = tail; tail = tail.prev; toRemove.prev = null; if (tail != null) { tail.next = null; } else { head = null; } }
This lets programs quickly and safely remove the last item from a list, making data handling faster and more reliable.
Think of a music playlist app where you want to remove the last song you added. Using a doubly linked list lets the app remove that song instantly without checking every song in the list.
Manual removal from the end is slow and error-prone.
Doubly linked lists keep track of the end for quick access.
Deleting from the end updates links both ways for safety.
Practice
Solution
Step 1: Identify the last node and its previous node
In a doubly linked list, the last node's previous node is the second last node.Step 2: Update the second last node's next pointer
Set the second last node's next pointer to null to remove the last node from the list.Final Answer:
The second last node becomes the new tail and its next pointer is set to null. -> Option DQuick Check:
Deleting last node updates tail's next to null [OK]
- Setting head to null instead of updating tail
- Not updating the previous node's next pointer
- Deleting the first node by mistake
dll?Solution
Step 1: Move tail pointer to previous node
Setdll.Tailtodll.Tail.Previousto remove the last node.Step 2: Set new tail's next pointer to null
Updatedll.Tail.Nextto null to mark the end of the list.Final Answer:
dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; -> Option BQuick Check:
Correct tail update syntax = dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; [OK]
- Updating head instead of tail
- Not setting tail's next to null
- Setting both head and tail to null unnecessarily
dll.AddLast(1); dll.AddLast(2); dll.AddLast(3); dll.DeleteFromEnd(); dll.PrintList();
Solution
Step 1: Add nodes 1, 2, 3 to the list
The list becomes 1 -> 2 -> 3 -> null.Step 2: Delete the last node (3)
After deletion, the list becomes 1 -> 2 -> null.Final Answer:
1 -> 2 -> null -> Option CQuick Check:
Last node removed, list ends at 2 [OK]
- Assuming list remains unchanged
- Deleting head instead of tail
- Printing list before deletion
if (dll.Tail != null) { dll.Tail = dll.Tail.Previous; dll.Tail.Next = null; }Solution
Step 1: Check deletion when only one node exists
If only one node exists,dll.Tail.Previousis null, sodll.Tailbecomes null.Step 2: Accessing
After settingdll.Tail.Nextcauses errordll.Tailto null, accessingdll.Tail.Nextthrows NullReferenceException.Final Answer:
NullReferenceException when list has only one node -> Option AQuick Check:
Accessing property on null tail causes error [OK]
- Not handling single-node list case
- Ignoring null checks before pointer access
- Assuming head updates automatically
Solution
Step 1: Initial list and first deletion
List is 10 <-> 20 <-> 30 <-> 40. Deleting last node removes 40, list becomes 10 <-> 20 <-> 30.Step 2: Second deletion removes 30
After second deletion, list becomes 10 <-> 20, and tail points to node with value 20.Final Answer:
10 <-> 20 -> null, Tail = 20 -> Option AQuick Check:
Two deletions remove last two nodes [OK]
- Deleting only one node instead of two
- Not updating tail after deletions
- Confusing head and tail pointers
Ready to prove it?
Pass the test and this pattern moves unknown
