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

Time Complexity: Reverse a Singly Linked List Iterative

O(n)

Understanding Time Complexity

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

Specifically, how many steps does it take to reverse all the nodes one by one?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Reverse a singly linked list iteratively
struct Node* reverseList(struct Node* head) {
    struct Node* prev = NULL;
    struct Node* curr = head;
    struct Node* next = NULL;
    while (curr != NULL) {
        next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}
    

This code reverses the links between nodes in a singly linked list 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 as many times as there are nodes in the list.

How Execution Grows With Input

Each node is visited once, so the total steps grow 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: Doubling the list size roughly doubles the work done.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows in a straight line with the number of nodes.

Common Mistake

[X] Wrong: "Reversing a linked list takes more than linear time because of pointer changes."

[OK] Correct: Each node is visited only once, and pointer changes happen in constant time per node, so total time is linear.

Interview Connect

Understanding this linear time reversal is a key skill that shows you can work with linked lists efficiently.

Self-Check

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

Practice

1. What is the main purpose of the iterative method to reverse a singly linked list?

easy

A. To delete all nodes from the linked list

B. To create a new linked list with reversed elements

C. To sort the linked list in ascending order

D. To change the direction of the links between nodes one by one

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes connected in one direction only, from head to tail.

2. Step 2: Purpose of iterative reversal

   The iterative method changes the direction of each node's pointer to point to the previous node, reversing the list.

3. Final Answer:

   To change the direction of the links between nodes one by one -> Option D

4. Quick Check:

   Reversing means changing links direction [OK]

Hint: Reversing means flipping links, not creating new nodes [OK]

Common Mistakes:

• Thinking reversal creates a new list instead of changing links
• Confusing reversal with sorting
• Assuming nodes are deleted during reversal

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

easy

A. struct Node *prev = NULL, *current = head, *next = NULL;

B. struct Node *prev = head, *current = NULL, *next = NULL;

C. struct Node *prev = NULL, *current = NULL, *next = head;

D. struct Node *prev = head, *current = head, *next = head;

Solution

1. Step 1: Identify starting pointers

   For reversal, prev starts as NULL because the new tail points to NULL, current starts at head to traverse.

2. Step 2: Correct initialization

   Next pointer is initialized as NULL and updated inside the loop; so initial next = NULL is correct.

3. Final Answer:

   struct Node *prev = NULL, *current = head, *next = NULL; -> Option A

4. Quick Check:

   Prev NULL, current head to start [OK]

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

Common Mistakes:

• Setting prev to head initially
• Setting current to NULL at start
• Initializing next pointer incorrectly

3. What will be the output linked list after running the iterative reverse function on the list: 1 -> 2 -> 3 -> NULL?

medium

A. 2 -> 3 -> 1 -> NULL

B. 3 -> 2 -> 1 -> NULL

C. NULL -> 3 -> 2 -> 1

D. 1 -> 2 -> 3 -> NULL

Solution

1. Step 1: Trace reversal process

   Initially: 1->2->3->NULL. After reversal, links point backward: 3->2->1->NULL.

2. Step 2: Confirm final head

   The last node (3) becomes the new head, so output is 3->2->1->NULL.

3. Final Answer:

   3 -> 2 -> 1 -> NULL -> Option B

4. Quick Check:

   Reverse order of nodes [OK]

Hint: Reverse means last becomes first [OK]

Common Mistakes:

• Assuming list stays the same after reversal
• Confusing NULL placement
• Mixing node order incorrectly

4. Identify the error in this iterative reversal code snippet for a singly linked list in C:

struct Node* reverse(struct Node* head) {
  struct Node* prev = head;
  struct Node* current = head;
  struct Node* next = NULL;
  while (current != NULL) {
    next = current->next;
    current->next = prev;
    prev = current;
    current = next;
  }
  return prev;
}

medium

A. prev should be initialized to NULL, not head

B. current should be initialized to NULL, not head

C. next pointer is not updated inside the loop

D. Function should return current, not prev

Solution

1. Step 1: Check pointer initialization

   Prev must start as NULL because the new tail node points to NULL, not head.

2. Step 2: Analyze loop pointer updates

   Current and next are correctly updated; only prev initialization is wrong.

3. Final Answer:

   prev should be initialized to NULL, not head -> Option A

4. Quick Check:

   Prev NULL to mark new tail [OK]

Hint: Prev must start NULL to avoid cycle [OK]

Common Mistakes:

• Starting prev at head causing cycle
• Not updating next inside loop
• Returning wrong pointer at end

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> NULL, write the final linked list after applying iterative reversal twice in a row.

hard

A. NULL

B. 40 -> 30 -> 20 -> 10 -> NULL

C. 10 -> 20 -> 30 -> 40 -> NULL

D. 20 -> 10 -> 40 -> 30 -> NULL

Solution

1. Step 1: Reverse the list once

   Original: 10->20->30->40->NULL. After first reversal: 40->30->20->10->NULL.

2. Step 2: Reverse the reversed list again

   Reversing 40->30->20->10->NULL returns it to original: 10->20->30->40->NULL.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> NULL -> Option C

4. Quick Check:

   Double reversal restores original list [OK]

Hint: Reversing twice returns original list [OK]

Common Mistakes:

• Assuming double reversal changes list order
• Confusing node values after reversal
• Forgetting NULL at list end