Delete Node by Value in DSA Python - Time & Space Complexity

Time Complexity: Delete Node by Value

O(n)

Understanding Time Complexity

When we delete a node by value from a linked list, we want to know how long it takes as the list grows.

We ask: How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def delete_node(head, value):
    if not head:
        return None
    if head.val == value:
        return head.next
    current = head
    while current.next and current.next.val != value:
        current = current.next
    if current.next:
        current.next = current.next.next
    return head

This code removes the first node with the given value from a singly linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing 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 gets longer, the time to find the node grows roughly in a straight line.

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

Pattern observation: The time grows directly with the number of nodes; doubling nodes roughly doubles the steps.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Deleting a node by value always takes constant time because we just remove it."

[OK] Correct: We must first find the node, which can take time proportional to the list length.

Interview Connect

Understanding this helps you explain how linked lists work and how operations scale, a key skill in many coding discussions.

Self-Check

"What if the list was doubly linked? How would the time complexity change when deleting a node by value?"

Practice

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

easy

A. Removes the first node that contains the given value

B. Adds a new node with the given value at the end

C. Finds the node with the given value without removing it

D. Reverses the linked list

Solution

1. Step 1: Understand the operation

   Deleting a node by value means removing the first node that matches the given value from the linked list.

2. Step 2: Compare options

   Only Removes the first node that contains the given value describes removing the first node with the given value. Others describe different operations.

3. Final Answer:

   Removes the first node that contains the given value -> Option A

4. Quick Check:

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

Hint: Delete node by value removes first matching node [OK]

Common Mistakes:

• Confusing deletion by value with adding a node
• Thinking it deletes all nodes with the value
• Mixing deletion by value with reversing the list

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

easy

A. def delete_node(head, value):
current = head
while current:
if current.val == value:
current = current.next
break
current = current.next

B. def delete_node(head, value):
prev = None
current = head
while current:
if current.val == value:
if prev:
prev.next = current.next
else:
head = current.next
break
prev = current
current = current.next
return head

C. def delete_node(head, value):
for node in head:
if node.val == value:
head.remove(node)
break
return head

D. def delete_node(head, value):
head = [x for x in head if x != value]
return head

Solution

1. Step 1: Analyze the deletion logic

   Deleting a node requires tracking the previous node to update its next pointer. def delete_node(head, value): prev = None current = head while current: if current.val == value: if prev: prev.next = current.next else: head = current.next break prev = current current = current.next return head correctly uses prev and current pointers and updates links properly.

2. Step 2: Check other options

   def delete_node(head, value): current = head while current: if current.val == value: current = current.next break current = current.next only moves current pointer without changing links. def delete_node(head, value): for node in head: if node.val == value: head.remove(node) break return head uses invalid iteration on linked list nodes. def delete_node(head, value): head = [x for x in head if x != value] return head treats linked list as a list, which is incorrect.

3. Final Answer:

   Option B code correctly deletes node by value in a singly linked list -> Option B

4. Quick Check:

   Proper pointer update = def delete_node(head, value): prev = None current = head while current: if current.val == value: if prev: prev.next = current.next else: head = current.next break prev = current current = current.next return head [OK]

Hint: Use prev and current pointers to update links [OK]

Common Mistakes:

• Not updating previous node's next pointer
• Trying to iterate linked list like a Python list
• Not returning updated head after deletion

3. What will be the output linked list after deleting value 3 from the list: 1 -> 3 -> 5 -> 3 -> 7 -> null?

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def delete_node(head, value):
    prev = None
    current = head
    while current:
        if current.val == value:
            if prev:
                prev.next = current.next
            else:
                head = current.next
            break
        prev = current
        current = current.next
    return head

# Create list
head = Node(1)
head.next = Node(3)
head.next.next = Node(5)
head.next.next.next = Node(3)
head.next.next.next.next = Node(7)

head = delete_node(head, 3)

# Print list
current = head
result = []
while current:
    result.append(current.val)
    current = current.next
print(result)

medium

A. [1, 5, 3, 7]

B. [1, 3, 5, 7]

C. [3, 5, 3, 7]

D. [1, 5, 7]

Solution

1. Step 1: Identify the first node with value 3

   The list is 1 -> 3 -> 5 -> 3 -> 7. The first node with value 3 is the second node.

2. Step 2: Delete the first 3 and link previous node to next

   After deletion, the list becomes 1 -> 5 -> 3 -> 7.

3. Final Answer:

   [1, 5, 3, 7] -> Option A

4. Quick Check:

   First 3 removed, rest unchanged = [1, 5, 3, 7] [OK]

Hint: Delete only first matching node, keep rest intact [OK]

Common Mistakes:

• Deleting all nodes with value 3 instead of first only
• Not updating head when first node matches
• Skipping nodes after deletion

4. The following code tries to delete a node by value from a linked list but causes an error. What is the error?

def delete_node(head, value):
    current = head
    while current.next:
        if current.next.val == value:
            current.next = current.next.next
            break
        current = current.next
    return head

medium

A. Fails when the value is not found

B. Fails when the list is empty

C. Fails when the node to delete is the last node

D. Fails when the node to delete is the head node

Solution

1. Step 1: Analyze deletion logic

   The code checks current.next.val to find the node to delete, so it cannot delete the head node if it matches the value.

2. Step 2: Check edge cases

   If the head node has the value, the code never deletes it because it only looks at current.next.

3. Final Answer:

   Fails when the node to delete is the head node -> Option D

4. Quick Check:

   Head node deletion not handled = Fails when the node to delete is the head node [OK]

Hint: Check if head node matches before loop [OK]

Common Mistakes:

• Not handling deletion of head node separately
• Assuming current.next always exists without check
• Ignoring empty list case

5. You have a singly linked list: 10 -> 20 -> 30 -> 40 -> 50 -> null. You want to delete the node with value 10 and then insert a new node with value 15 after the node with value 20. What will be the final linked list? Choose the correct sequence of operations and final list. Options: A) Delete 10, insert 15 after 20: 20 -> 15 -> 30 -> 40 -> 50 B) Delete 10, insert 15 after 20: 10 -> 20 -> 15 -> 30 -> 40 -> 50 C) Delete 10, insert 15 after 20: 20 -> 30 -> 15 -> 40 -> 50 D) Delete 10, insert 15 after 20: 15 -> 20 -> 30 -> 40 -> 50

hard

A. 20 -> 30 -> 15 -> 40 -> 50

B. 10 -> 20 -> 15 -> 30 -> 40 -> 50

C. 20 -> 15 -> 30 -> 40 -> 50

D. 15 -> 20 -> 30 -> 40 -> 50

Solution

1. Step 1: Delete node with value 10

   Since 10 is the head, deleting it makes 20 the new head. List becomes 20 -> 30 -> 40 -> 50.

2. Step 2: Insert node with value 15 after node with value 20

   Find node 20, insert 15 after it. List becomes 20 -> 15 -> 30 -> 40 -> 50.

3. Final Answer:

   20 -> 15 -> 30 -> 40 -> 50 -> Option C

4. Quick Check:

   Delete head 10, insert 15 after 20 = 20 -> 15 -> 30 -> 40 -> 50 [OK]

Hint: Delete head carefully, then insert after target node [OK]

Common Mistakes:

• Not updating head after deleting first node
• Inserting new node in wrong position
• Confusing node values and positions