Reverse a Singly Linked List Iterative in DSA Python - Time & Space Complexity

Time Complexity: Reverse a Singly Linked List Iterative

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a singly linked list changes as the list gets bigger.

Specifically, how does the number of steps grow when we reverse the list one node at a time?

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 reverse_linked_list(head):
    prev = None
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

This code reverses a singly linked list by changing the direction of the links one by one until the entire list is reversed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node, so n times for n nodes.

How Execution Grows With Input

As the list gets longer, the number of steps grows directly with the number of nodes.

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

Pattern observation: The steps increase in a straight line as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly in proportion to the number of nodes.

Common Mistake

[X] Wrong: "Reversing the list takes constant time because we just change pointers."

[OK] Correct: Even though each pointer change is quick, we must do it for every node, so the total time grows with the list size.

Interview Connect

Understanding this time complexity helps you explain why reversing a list is efficient and what to expect as input grows, a useful skill in many coding challenges.

Self-Check

"What if we used recursion instead of iteration to reverse the list? How would the time complexity change?"

Practice

1. What is the main purpose of reversing a singly linked list iteratively?

easy

A. To remove duplicate nodes from the list

B. To sort the list in ascending order

C. To change the direction of all links so the list order is reversed

D. To find the middle node of the list

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes pointing only to the next node, so reversing means changing these pointers to point backwards.

2. Step 2: Purpose of iterative reversal

   The iterative method changes each node's next pointer to the previous node, effectively reversing the list order.

3. Final Answer:

   To change the direction of all links so the list order is reversed -> Option C

4. Quick Check:

   Reversing means flipping links direction [OK]

Hint: Reversing flips all next pointers to previous nodes [OK]

Common Mistakes:

• Confusing reversal with sorting
• Thinking reversal deletes nodes
• Assuming reversal finds middle node

2. Which of the following is the correct way to initialize pointers for iterative reversal of a singly linked list in Python?

easy

A. prev = head; current = head.next

B. prev = null; current = head

C. prev = head; current = null

D. prev = null; current = null

Solution

1. Step 1: Understand pointer roles

   Prev starts as null because the new tail will point to null; current starts at head to traverse the list.

2. Step 2: Correct initialization

   Setting prev = null and current = head allows iteration from the start with no previous node initially.

3. Final Answer:

   prev = null; current = head -> Option B

4. Quick Check:

   Prev null, current at head to start [OK]

Hint: Prev starts null, current starts at head [OK]

Common Mistakes:

• Setting prev to head initially
• Starting current as null
• Both pointers as null

3. What will be the output linked list after running the iterative reversal code on this list: 1 -> 2 -> 3 -> null?

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null -> 3 -> 2 -> 1

D. 2 -> 3 -> 1 -> null

Solution

1. Step 1: Trace reversal process

   Initially: prev=null, current=1. After first iteration, 1 points to null. Then 2 points to 1, then 3 points to 2.

2. Step 2: Final reversed list

   The list becomes 3 -> 2 -> 1 -> null after all nodes reverse their next pointers.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list order is 3, 2, 1 [OK]

Hint: Last node becomes new head after reversal [OK]

Common Mistakes:

• Not updating pointers correctly
• Confusing original and reversed order
• Including null as a node

4. Identify the error in this iterative reversal code snippet:

def reverse(head):
    prev = head
    current = null
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

medium

A. prev should start as null, current as head

B. The loop condition should be 'while prev' instead of 'while current'

C. current.next assignment is reversed, should be 'prev.next = current'

D. Function should return current instead of prev

Solution

1. Step 1: Check pointer initialization

   Prev should be null initially because the new tail points to null; current should start at head to traverse.

2. Step 2: Loop condition and pointer updates

   Loop runs while current is not null; current.next is set to prev correctly; returning prev gives new head.

3. Final Answer:

   prev should start as null, current as head -> Option A

4. Quick Check:

   Prev null, current head to start loop [OK]

Hint: Prev null, current head to avoid skipping nodes [OK]

Common Mistakes:

• Starting prev at head causes cycle
• Loop condition incorrect
• Returning wrong pointer

5. Given a singly linked list with nodes containing values [10, 20, 30, 40, 50], which iterative reversal step correctly updates the pointers after processing the first two nodes?

hard

A. After first node: 10 -> 20; After second node: 20 -> null

B. After first node: 10 -> 20 -> null; After second node: 20 -> 10 -> null

C. After first node: null -> 10; After second node: 10 -> 20

D. After first node: 10 -> null; After second node: 20 -> 10 -> null

Solution

1. Step 1: Reverse first node pointer

   Initially, prev=null, current=10. After first iteration, 10.next = null, so list is 10 -> null.

2. Step 2: Reverse second node pointer

   Next, current=20. Set 20.next = prev (which is 10), so list is 20 -> 10 -> null.

3. Final Answer:

   After first node: 10 -> null; After second node: 20 -> 10 -> null -> Option D

4. Quick Check:

   Each node points back to previous [OK]

Hint: Each step points current node to previous node [OK]

Common Mistakes:

• Not setting first node's next to null
• Linking nodes forward instead of backward
• Confusing order of pointer updates