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DSA C#programming~5 mins

Delete Node at End in DSA C# - Time & Space Complexity

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Time Complexity: Delete Node at End
O(n)
Understanding Time Complexity

We want to understand how long it takes to remove the last item from a linked list.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


public void DeleteAtEnd()
{
    if (head == null) return;
    if (head.next == null) {
        head = null;
        return;
    }
    Node current = head;
    while (current.next.next != null) {
        current = current.next;
    }
    current.next = null;
}
    

This code removes the last node from a singly linked list by walking to the second last node and cutting off the last one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: The while loop that moves through nodes to find the second last node.
  • How many times: It runs once for each node except the last two, so roughly n-2 times for a list of size n.
How Execution Grows With Input

As the list grows, the loop runs more times because it must reach near the end.

Input Size (n)Approx. Operations
10About 8 times through the loop
100About 98 times through the loop
1000About 998 times through the loop

Pattern observation: The number of steps grows roughly the same as the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the last node grows linearly with the list size.

Common Mistake

[X] Wrong: "Deleting the last node is always fast and takes constant time."

[OK] Correct: Because we must find the node before the last one, which takes time proportional to the list length.

Interview Connect

Understanding this helps you explain how linked lists work and why some operations take longer as the list grows.

Self-Check

"What if the list had a tail pointer that points directly to the last node? How would the time complexity change?"

Practice

(1/5)
1. What does the operation Delete Node at End do in a singly linked list?
easy
A. Adds a new node at the end of the list
B. Removes the first node from the list
C. Removes the last node from the list
D. Adds a new node at the beginning of the list

Solution

  1. Step 1: Understand the operation meaning

    Deleting a node at the end means removing the last node in the linked list.
  2. Step 2: Compare options with operation

    Only Removes the last node from the list matches removing the last node; others add or remove nodes elsewhere.
  3. Final Answer:

    Removes the last node from the list -> Option C
  4. Quick Check:

    Delete Node at End = Remove last node [OK]
Hint: Deleting at end means removing the last node [OK]
Common Mistakes:
  • Confusing delete at end with delete at beginning
  • Thinking it adds a node instead of removing
  • Assuming it deletes a middle node
2. Which of the following is the correct syntax to delete the last node in a singly linked list in C#?
easy
A. Create a new node and assign it to last node
B. Set head to null directly
C. Set last node's next to head
D. Traverse to second last node, set its next to null

Solution

  1. Step 1: Identify correct deletion method

    To delete last node, we must reach second last node and set its next pointer to null.
  2. Step 2: Check options for this logic

    Traverse to second last node, set its next to null matches this logic; others do unrelated actions.
  3. Final Answer:

    Traverse to second last node, set its next to null -> Option D
  4. Quick Check:

    Delete last node = second last node next = null [OK]
Hint: Set second last node's next to null to delete last [OK]
Common Mistakes:
  • Setting head to null deletes whole list
  • Trying to set last node's next to head causes cycle
  • Creating new node does not delete existing node
3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what will be the list after deleting the node at the end?
medium
A. 10 -> 20 -> 30 -> 40 -> null
B. 10 -> 20 -> 30 -> null
C. 20 -> 30 -> 40 -> null
D. 10 -> 20 -> null

Solution

  1. Step 1: Identify last node in list

    The last node is 40 in the list 10 -> 20 -> 30 -> 40 -> null.
  2. Step 2: Remove last node and update list

    Deleting last node removes 40, so list becomes 10 -> 20 -> 30 -> null.
  3. Final Answer:

    10 -> 20 -> 30 -> null -> Option B
  4. Quick Check:

    Last node 40 removed = 10->20->30->null [OK]
Hint: Remove last value to get new list [OK]
Common Mistakes:
  • Not removing last node, list unchanged
  • Removing first node instead
  • Removing more than one node accidentally
4. Consider this code snippet to delete the last node of a singly linked list:
Node current = head;
while(current.next.next != null) {
    current = current.next;
}
current.next = null;
What error can occur when this code runs on a list with only one node?
medium
A. NullReferenceException because current.next.next is invalid
B. No error, code works fine
C. Infinite loop because condition never false
D. Deletes the head node incorrectly

Solution

  1. Step 1: Analyze condition for single node list

    If list has one node, current.next is null, so current.next.next causes error.
  2. Step 2: Identify exception type

    Accessing next of null causes NullReferenceException in C#.
  3. Final Answer:

    NullReferenceException because current.next.next is invalid -> Option A
  4. Quick Check:

    Single node list causes null.next access error [OK]
Hint: Check for null before accessing next.next [OK]
Common Mistakes:
  • Ignoring null check for single node
  • Assuming loop runs safely always
  • Thinking it deletes head node
5. You have a singly linked list with nodes: 5 -> 15 -> 25 -> 35 -> 45 -> null. Write a method to delete the last node and return the updated list. Which of the following code snippets correctly implements this?
hard
A. if (head == null) return null; if (head.next == null) return null; Node current = head; while (current.next.next != null) { current = current.next; } current.next = null; return head;
B. if (head == null) return null; Node current = head; while (current.next != null) { current = current.next; } current = null; return head;
C. if (head == null) return null; Node current = head; while (current.next.next != null) { current = current.next; } current.next = current; return head;
D. if (head == null) return null; head = head.next; return head;

Solution

  1. Step 1: Handle empty and single node cases

    If list is empty or has one node, deleting last node returns null (empty list).
  2. Step 2: Traverse to second last node and delete last

    Loop until current.next.next is null, then set current.next to null to remove last node.
  3. Step 3: Return updated head

    Return head pointer to updated list.
  4. Final Answer:

    Code snippet D correctly deletes last node and returns updated list -> Option A
  5. Quick Check:

    Check null, traverse to second last, set next null [OK]
Hint: Check list size, then remove last by second last next = null [OK]
Common Mistakes:
  • Setting current to null does not remove last node
  • Assigning current.next to current creates cycle
  • Deleting head instead of last node
Recommended

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10 questions~5 mineasy hard