Pop Using Linked List Node in DSA Python - Time & Space Complexity

Time Complexity: Pop Using Linked List Node

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 as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

def pop(head):
    if head is None:
        return None
    if head.next is None:
        return None  # List becomes empty after pop
    current = head
    while current.next.next is not None:
        current = current.next
    current.next = None
    return head

This code removes the last node from a singly linked list by walking through the list to find the second last node.

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 until the second last one, so about n-2 times for a list of size n.

How Execution Grows With Input

As the list gets longer, the loop runs more times, almost once per node.

Input Size (n)	Approx. Operations
10	About 8 steps
100	About 98 steps
1000	About 998 steps

Pattern observation: The number of steps grows roughly in a straight line with the size of the list.

Final Time Complexity

Time Complexity: O(n)

This means the time to pop the last node grows directly with the number of nodes in the list.

Common Mistake

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

[OK] Correct: Because we must find the second last node by walking through the list, which takes longer as the list grows.

Interview Connect

Understanding this helps you explain why some linked list operations are slower than others and shows you can analyze code carefully.

Self-Check

"What if the linked list kept a pointer to the last node? How would the time complexity of pop change?"

Practice

1. What does the pop operation do in a linked list when using nodes?

easy

A. Removes the last node and returns its value

B. Adds a new node at the beginning

C. Searches for a node with a specific value

D. Counts the total number of nodes

Solution

1. Step 1: Understand pop operation in linked list

   Pop means removing the last element from the list and returning its value.

2. Step 2: Identify pop behavior with nodes

   In a linked list, pop removes the tail node and returns its data.

3. Final Answer:

   Removes the last node and returns its value -> Option A

4. Quick Check:

   Pop = Remove last node [OK]

Hint: Pop always removes the last node in a linked list [OK]

Common Mistakes:

• Confusing pop with push (adding nodes)
• Thinking pop removes the first node
• Assuming pop searches for a value

2. Which of the following is the correct syntax to pop a node from a singly linked list in Python?

easy

A. def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data

B. def pop(self):\n popped = self.head\n self.head = self.head.next\n return popped.data

C. def pop(self):\n return self.head.data

D. def pop(self):\n self.head = None

Solution

1. Step 1: Check pop logic for singly linked list

   Pop removes the last node, so we must traverse to the second last node.

2. Step 2: Verify syntax correctness

   def pop(self):\n if not self.head:\n return None\n current = self.head\n while current.next.next:\n current = current.next\n popped = current.next\n current.next = None\n return popped.data correctly traverses until current.next.next is None, then removes last node and returns its data.

3. Final Answer:

   Code that traverses to second last node -> Option A

4. Quick Check:

   Traverse to second last node, remove last [OK]

Hint: Pop needs traversal to second last node before removal [OK]

Common Mistakes:

• Removing head instead of last node
• Not checking for empty list
• Returning node instead of node data

3. What will be the output after popping a node from the linked list 1 -> 2 -> 3 -> null using the pop method below?

def pop(self):
    if not self.head:
        return None
    if not self.head.next:
        val = self.head.data
        self.head = None
        return val
    current = self.head
    while current.next.next:
        current = current.next
    val = current.next.data
    current.next = None
    return val

medium

A. 2 -> 3 -> null

B. 1 -> 3 -> null

C. 1 -> 2 -> null

D. 3 -> null

Solution

1. Step 1: Identify the node popped

   Pop removes the last node, which is 3 in the list 1 -> 2 -> 3 -> null.

2. Step 2: Determine list after pop

   After removing 3, the list becomes 1 -> 2 -> null.

3. Final Answer:

   1 -> 2 -> null -> Option C

4. Quick Check:

   Last node removed, list ends at 2 [OK]

Hint: Pop removes last node, list shortens by one [OK]

Common Mistakes:

• Removing first node instead of last
• Not updating next pointer to None
• Returning wrong list after pop

4. Find the error in this pop method for a singly linked list with at least two nodes:

def pop(self):
    current = self.head
    while current.next:
        current = current.next
    val = current.data
    current = None
    return val

medium

A. It does not handle empty list case

B. It does not update the previous node's next pointer

C. It returns None instead of node data

D. It removes the head instead of last node

Solution

1. Step 1: Analyze traversal and removal

   The code moves current to last node but does not track previous node.

2. Step 2: Check pointer update

   Setting current = None only changes local variable, does not update previous node's next pointer to None, so last node remains linked.

3. Final Answer:

   It does not update the previous node's next pointer -> Option B

4. Quick Check:

   Must update previous.next to None to remove last node [OK]

Hint: Always update previous node's next to None when popping last node [OK]

Common Mistakes:

• Not handling empty list
• Not tracking previous node
• Setting local variable to None instead of pointer

5. Given a linked list with nodes 10 -> 20 -> 30 -> 40 -> null, write the sequence of node values after two consecutive pop operations using a correct pop method that removes the last node each time.

hard

A. 10 -> 30 -> null

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

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

D. 10 -> 20 -> null

Solution

1. Step 1: First pop removes last node 40

   List changes from 10 -> 20 -> 30 -> 40 -> null to 10 -> 20 -> 30 -> null.

2. Step 2: Second pop removes new last node 30

   List changes from 10 -> 20 -> 30 -> null to 10 -> 20 -> null.

3. Final Answer:

   10 -> 20 -> null -> Option D

4. Quick Check:

   Two pops remove last two nodes [OK]

Hint: Each pop removes last node, shorten list stepwise [OK]

Common Mistakes:

• Removing nodes from start instead of end
• Not updating list after first pop
• Confusing node values after pops