Delete Node by Value in DSA C# - Time & Space Complexity

Time Complexity: Delete Node by Value

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove a node by its value 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 DeleteNode(int value) {
    Node current = head;
    Node previous = null;
    while (current != null) {
        if (current.Data == value) {
            if (previous == null) {
                head = current.Next;
            } else {
                previous.Next = current.Next;
            }
            return;
        }
        previous = current;
        current = current.Next;
    }
}
    

This code searches the linked list for the first node with the given value and removes it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through the linked list nodes one by one.
• How many times: Up to n times, where n is the number of nodes in the list.

How Execution Grows With Input

As the list grows, the time to find the node grows roughly in a straight line.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

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

Final Time Complexity

Time Complexity: O(n)

This means the time to delete a node grows in a straight line as the list gets longer.

Common Mistake

[X] Wrong: "Deleting a node by value always takes the same time no matter the list size."

[OK] Correct: Because you may need to look through many nodes before finding the value, so time grows with list length.

Interview Connect

Understanding this helps you explain how searching and deleting in linked lists works, a key skill for many coding tasks.

Self-Check

"What if the list was doubly linked? How would the time complexity change?"

Practice

1. What does the operation Delete Node by Value do in a linked list?

easy

A. Adds a new node with the given value at the end of the list

B. Finds the node with the given value and returns it without removing

C. Removes the first node that contains the given value from the list

D. Reverses the linked list

Solution

1. Step 1: Understand the operation

   Deleting a node by value means searching for the first node that matches the value and removing it from the linked list.

2. Step 2: Compare options

   Only Removes the first node that contains the given value from the list describes removing the first node with the given value. Others describe adding, finding without removing, or reversing, which are different operations.

3. Final Answer:

   Removes the first node that contains the given value from the list -> Option C

4. Quick Check:

   Delete Node by Value = Remove first matching node [OK]

Hint: Deleting by value removes first matching node only [OK]

Common Mistakes:

• Confusing deletion by value with adding a node
• Thinking it removes all nodes with that value
• Assuming it reverses the list

2. Which of the following is the correct C# syntax to delete a node by value from a singly linked list?

easy

A. void DeleteNodeByValue(int value) { /* code */ }

B. public void DeleteNode(int value) { /* code */ }

C. public int DeleteNode(int value) { /* code */ }

D. DeleteNode(int value) => { /* code */ }

Solution

1. Step 1: Check method signature conventions in C#

   In C#, methods usually have an access modifier like 'public' and a return type. Deleting a node usually returns void.

2. Step 2: Analyze options

   public void DeleteNode(int value) { /* code */ } uses 'public void DeleteNode(int value)', which is correct. void DeleteNodeByValue(int value) { /* code */ } misses access modifier. public int DeleteNode(int value) { /* code */ } returns int which is uncommon here. DeleteNode(int value) => { /* code */ } uses invalid lambda syntax for method.

3. Final Answer:

   public void DeleteNode(int value) { /* code */ } -> Option B

4. Quick Check:

   Correct method signature = public void DeleteNode(int value) { /* code */ } [OK]

Hint: Use public void method with parameter for deletion [OK]

Common Mistakes:

• Omitting access modifier
• Using wrong return type
• Incorrect method syntax

3. Given the linked list: 10 -> 20 -> 30 -> 40 -> null, what is the list after deleting the node with value 30 using the DeleteNodeByValue method?

medium

A. 10 -> 20 -> 40 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 20 -> 30 -> 40 -> null

D. 10 -> 30 -> 40 -> null

Solution

1. Step 1: Identify the node to delete

   The node with value 30 is the third node in the list.

2. Step 2: Remove the node and link previous to next

   After deleting 30, the node with value 20 points directly to 40, so the list becomes 10 -> 20 -> 40 -> null.

3. Final Answer:

   10 -> 20 -> 40 -> null -> Option A

4. Quick Check:

   Deleted 30, list skips it [OK]

Hint: Remove node and link previous to next node [OK]

Common Mistakes:

• Not updating links after deletion
• Deleting wrong node
• Leaving deleted node in list

4. Consider this code snippet to delete a node by value in a singly linked list:

public void DeleteNode(int value) {
  Node current = head;
  Node previous = null;
  while (current != null && current.data != value) {
    previous = current;
    current = current.next;
  }
  if (current == null) return;
  previous.next = current.next;
}

What error will occur when deleting the head node?

medium

A. Compilation error due to missing return type

B. No error, deletion works fine

C. Infinite loop because current never changes

D. NullReferenceException because previous is null when deleting head

Solution

1. Step 1: Analyze deletion of head node

   If the node to delete is head, previous is still null because loop never runs.

2. Step 2: Check assignment previous.next = current.next

   Since previous is null, trying to access previous.next causes NullReferenceException.

3. Final Answer:

   NullReferenceException because previous is null when deleting head -> Option D

4. Quick Check:

   Deleting head needs special case to avoid null error [OK]

Hint: Handle head deletion separately to avoid null errors [OK]

Common Mistakes:

• Not checking if node to delete is head
• Assuming previous is never null
• Ignoring null pointer exceptions

5. You have a singly linked list: 5 -> 10 -> 15 -> 20 -> 25 -> null. You want to delete the node with value 5 (head), then delete the node with value 20. What will the list look like after these two deletions?

hard

A. 10 -> 15 -> 25 -> null

B. 5 -> 10 -> 15 -> 25 -> null

C. 10 -> 15 -> 20 -> 25 -> null

D. 15 -> 25 -> null

Solution

1. Step 1: Delete head node with value 5

   Removing head means head moves to next node, so list becomes 10 -> 15 -> 20 -> 25 -> null.

2. Step 2: Delete node with value 20

   Find node 20 and remove it by linking 15 to 25, resulting in 10 -> 15 -> 25 -> null.

3. Final Answer:

   10 -> 15 -> 25 -> null -> Option A

4. Quick Check:

   Deleted head and middle node correctly [OK]

Hint: Delete head first, then link previous to next for others [OK]

Common Mistakes:

• Not updating head after deleting first node
• Not linking nodes correctly after deletion
• Deleting wrong nodes