Delete Node by Value in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does 'Delete Node by Value' mean in a linked list?

It means finding the first node that contains a specific value and removing it from the linked list, adjusting links so the list stays connected.

beginner

Why do we need to handle the head node differently when deleting by value?

Because the head node is the start of the list, if it contains the value to delete, we must update the head pointer to the next node to keep the list valid.

beginner

What happens if the value to delete is not found in the linked list?

No nodes are removed, and the linked list remains unchanged.

intermediate

In the deletion process, why do we keep track of the previous node?

Because to remove a node, we need to change the previous node's next pointer to skip the node being deleted.

intermediate

What is the time complexity of deleting a node by value in a singly linked list?

O(n), where n is the number of nodes, because we may need to check each node until we find the value or reach the end.

What should you do if the node to delete is the head node in a singly linked list?

A Do nothing

B Delete the entire list

C Update the head pointer to the next node

D Swap the head with the last node

The head pointer must be updated to the next node to remove the current head.

If the value to delete is not found, what happens to the linked list?

A It remains unchanged

B The head node is deleted

C The last node is deleted

D It becomes empty

No deletion occurs if the value is not found, so the list stays the same.

Why do we need to keep track of the previous node when deleting a node by value?

A To update the previous node's next pointer

B To print the list

C To find the last node

D To reverse the list

We update the previous node's next pointer to skip the node being deleted.

What is the worst-case time complexity of deleting a node by value in a singly linked list?

A O(1)

B O(n)

C O(log n)

D O(n^2)

We may need to traverse the entire list to find the node, so it is O(n).

Which of these is NOT a step in deleting a node by value?

A Handle the case when the node is the head

B Update the previous node's next pointer

C Find the node with the value

D Create a new node with the value

Creating a new node is not part of deletion; deletion removes nodes.

Explain step-by-step how to delete a node by value in a singly linked list.

Think about how to keep the list connected after removing the node.

What special cases must you consider when deleting a node by value in a linked list?

Consider the start of the list and empty list scenarios.

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