Delete Node at Specific Position in DSA Python - Build from Scratch

Delete Node at Specific Position

📖 Scenario: Imagine you have a chain of friends holding hands in a line. Sometimes, one friend needs to leave the line from a specific spot. We want to help by removing that friend without breaking the chain.

🎯 Goal: You will build a simple linked list and write code to remove a friend (node) from a specific position in the chain.

📋 What You'll Learn

Create a linked list with exact nodes: 10, 20, 30, 40, 50

Create a variable called position to hold the position of the node to delete

Write code to delete the node at the given position in the linked list

Print the linked list after deletion in the format: 10 -> 20 -> 40 -> 50 -> None

💡 Why This Matters

🌍 Real World

Linked lists are used in many software systems to manage ordered data where items can be added or removed efficiently, like music playlists or undo history in apps.

💼 Career

Understanding linked lists and how to manipulate them is a fundamental skill for software developers, especially when working with low-level data structures or optimizing performance.

1

Create the linked list

Create a linked list with nodes containing these exact values in order: 10, 20, 30, 40, 50. Use a class called Node with attributes data and next. Create a variable called head that points to the first node.

DSA Python

# Define Node class and create linked list with nodes 10, 20, 30, 40, 50
# Your code here

Hint

Start by defining the Node class with data and next. Then create each node and link them by setting the next attribute.

2

Set the position to delete

Create a variable called position and set it to 3. This means you want to delete the node at position 3 in the linked list (counting starts at 1).

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

# Create variable position and set it to 3
# Your code here

Hint

Just create a variable named position and assign the value 3 to it.

3

Delete the node at the given position

Write code to delete the node at the position stored in position from the linked list starting at head. If position is 1, update head to the next node. Otherwise, find the node before the position and change its next to skip the node at position.

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

position = 3

# Delete the node at position
# Your code here

Hint

Check if position is 1 to update head. Otherwise, move to the node before the one to delete and change its next to skip the deleted node.

4

Print the linked list after deletion

Write code to print the linked list starting from head in the format: 10 -> 20 -> 40 -> 50 -> None. Use a variable called current to traverse the list and print each node's data followed by ->. End with None.

DSA Python

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

head = Node(10)
second = Node(20)
third = Node(30)
fourth = Node(40)
fifth = Node(50)

head.next = second
second.next = third
third.next = fourth
fourth.next = fifth

position = 3

if position == 1:
    head = head.next
else:
    current = head
    count = 1
    while current is not None and count < position - 1:
        current = current.next
        count += 1
    if current is not None and current.next is not None:
        current.next = current.next.next

# Print the linked list
# Your code here

Hint

Use a loop to go through each node starting at head. Print the data and an arrow until you reach the end, then print None.

Practice

1. What happens when you delete a node at position 0 in a singly linked list?

easy

A. The last node is removed

B. The list remains unchanged

C. The head node is removed and the next node becomes the new head

D. All nodes are deleted

Solution

1. Step 1: Understand position 0 in linked list

   Position 0 refers to the head node of the list.

2. Step 2: Effect of deleting head node

   Deleting the head means the next node becomes the new head.

3. Final Answer:

   The head node is removed and the next node becomes the new head -> Option C

4. Quick Check:

   Deleting position 0 removes head [OK]

Hint: Deleting position 0 always removes the head node [OK]

Common Mistakes:

• Thinking last node is deleted at position 0
• Assuming list stays same after deletion
• Deleting all nodes instead of one

2. Which of the following is the correct syntax to delete a node at position pos in a singly linked list in Python?

easy

A. head = head.next if pos == 1 else head

B. if pos == 0: head = head.next

C. head.next = head if pos == 0 else head.next

D. head = head.prev if pos == 0 else head.next

Solution

1. Step 1: Check deletion at position 0

   Deleting at position 0 means moving head to next node.

2. Step 2: Verify syntax correctness

   if pos == 0: head = head.next correctly assigns head to head.next when pos is 0.

3. Final Answer:

   if pos == 0: head = head.next -> Option B

4. Quick Check:

   Correct syntax for deleting head node [OK]

Hint: Deleting head means head = head.next [OK]

Common Mistakes:

• Using head.prev which doesn't exist in singly linked list
• Wrong condition for position check
• Assigning head incorrectly

3. What will be the linked list after deleting the node at position 2 from the list: 10 -> 20 -> 30 -> 40 -> null?

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

def delete_node(head, pos):
    if pos == 0:
        return head.next
    current = head
    for _ in range(pos - 1):
        current = current.next
    current.next = current.next.next
    return head

head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head = delete_node(head, 2)

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

medium

A. [10, 20, 40]

B. [10, 30, 40]

C. [20, 30, 40]

D. [10, 20, 30, 40]

Solution

1. Step 1: Identify node at position 2

   Position 2 node has value 30 in the list 10 -> 20 -> 30 -> 40.

2. Step 2: Delete node at position 2

   Node 20's next pointer is updated to node 40, skipping node 30.

3. Final Answer:

   [10, 20, 40] -> Option A

4. Quick Check:

   Deleted node 30 at position 2 [OK]

Hint: Skip node at position by linking previous to next [OK]

Common Mistakes:

• Deleting wrong node due to off-by-one error
• Not updating next pointer correctly
• Returning wrong head

4. Find the error in the following code snippet that deletes a node at position pos in a singly linked list:

def delete_node(head, pos):
    if pos == 0:
        head = head.next
    current = head
    for _ in range(pos - 1):
        current = current.next
    current.next = current.next.next
    return head

medium

A. No error, code is correct

B. Loop runs one time too many

C. current.next.next may cause AttributeError if current.next is None

D. Missing return after deleting head node

Solution

1. Step 1: Check deletion at position 0

   When pos is 0, head is updated but not returned immediately.

2. Step 2: Effect of missing return

   Code continues and tries to access current.next causing errors or wrong behavior.

3. Final Answer:

   Missing return after deleting head node -> Option D

4. Quick Check:

   Return immediately after deleting head [OK]

Hint: Return immediately after deleting head node [OK]

Common Mistakes:

• Not returning head after deleting first node
• Ignoring edge cases when list is empty
• Assuming current.next always exists

5. Given a singly linked list 5 -> 10 -> 15 -> 20 -> 25 -> null, write the resulting list after deleting nodes at positions 0, 2, and then 1 in that order. What is the final list?

hard

A. [10, 25]

B. [10, 15, 25]

C. [10, 20]

D. [5, 15, 25]

Solution

1. Step 1: Delete node at position 0

   Deleting head (5), list becomes 10 -> 15 -> 20 -> 25.

2. Step 2: Delete node at position 2

   Node at position 2 is 20, deleting it results in 10 -> 15 -> 25.

3. Step 3: Delete node at position 1

   Node at position 1 is 15, deleting it results in 10 -> 25.

4. Final Answer:

   [10, 25] -> Option A

5. Quick Check:

   Sequential deletions update list correctly [OK]

Hint: Delete nodes stepwise, update positions after each deletion [OK]

Common Mistakes:

• Not updating list after each deletion
• Confusing positions after deletions
• Deleting wrong nodes due to static indexing