Reverse a Singly Linked List Iterative in DSA Typescript - 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 linked list changes as the list gets longer.

How does the number of steps grow when the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function reverseList(head: ListNode | null): ListNode | null {
  let prev = null;
  let current = head;
  while (current !== null) {
    const next = current.next;
    current.next = prev;
    prev = current;
    current = next;
  }
  return prev;
}
    

This code reverses a singly linked list by changing the direction of the links one by one.

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 the list length.

How Execution Grows With Input

Each node is visited once, so the 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: The time grows in a straight line as the list gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly 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 changing pointers is a simple step done in constant time per node.

Interview Connect

Understanding this helps you explain how to efficiently reverse lists, a common task in coding problems and real applications.

Self-Check

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

Practice

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

easy

A. To sort the linked list in ascending order

B. To change the direction of the links between nodes without using recursion

C. To delete all nodes from the linked list

D. To find the middle node of the linked list

Solution

1. Step 1: Understand the operation of reversing

   Reversing a singly linked list means changing the direction of the next pointers so the last node becomes the first.

2. Step 2: Identify iterative approach

   The iterative method changes links step-by-step using a loop without recursion.

3. Final Answer:

   To change the direction of the links between nodes without using recursion -> Option B

4. Quick Check:

   Reversing iteratively = changing links direction without recursion [OK]

Hint: Reversing means flipping links, not sorting or deleting [OK]

Common Mistakes:

• Confusing reversing with sorting
• Thinking recursion is required
• Assuming nodes are deleted

2. Which of the following TypeScript code snippets correctly initializes pointers for iterative reversal of a singly linked list?

easy

A. let prev = null; let current = head;

B. let prev = head; let current = null;

C. let prev = head; let current = head.next;

D. let prev = null; let current = null;

Solution

1. Step 1: Understand pointer roles

   Prev starts as null because the new tail points to null; current starts at head to traverse.

2. Step 2: Match initialization

   Only let prev = null; let current = head; sets prev to null and current to head correctly for iteration.

3. Final Answer:

   let prev = null; let current = head; -> Option A

4. Quick Check:

   Prev null, current head = correct start [OK]

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

Common Mistakes:

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

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

function reverseList(head: ListNode | null): ListNode | null {
  let prev = null;
  let current = head;
  while (current !== null) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  return prev;
}

medium

A. null

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

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

D. 1 -> null

Solution

1. Step 1: Trace the reversal loop

   Initially prev=null, current=1. Each iteration reverses current.next to prev and moves pointers forward.

2. Step 2: Final reversed list

   After loop, prev points to node 3, which links to 2, then 1, then null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option C

4. Quick Check:

   Reversed list = 3 -> 2 -> 1 -> null [OK]

Hint: Follow pointers step-by-step to see reversed links [OK]

Common Mistakes:

• Returning head instead of prev
• Not updating current correctly
• Confusing next pointer updates

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

function reverseList(head: ListNode | null): ListNode | null {
  let prev = head;
  let current = head.next;
  while (current !== null) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  return prev;
}

medium

A. current should start at head, not head.next

B. The function should return current instead of prev

C. nextTemp should be assigned after updating current.next

D. prev should be initialized to null, not head

Solution

1. Step 1: Check pointer initialization

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

2. Step 2: Understand impact of wrong initialization

   Starting prev at head causes a cycle or incorrect reversal.

3. Final Answer:

   prev should be initialized to null, not head -> Option D

4. Quick Check:

   Prev must be null at start [OK]

Hint: Prev always null at start to avoid cycles [OK]

Common Mistakes:

• Starting prev at head causing loops
• Starting current at wrong node
• Returning wrong pointer

5. You have a singly linked list with nodes containing values: 10 -> 20 -> 30 -> 40 -> null. After reversing it iteratively, you want to insert a new node with value 25 between 20 and 30 in the reversed list. What will be the correct order of nodes after insertion?

hard

A. 40 -> 30 -> 25 -> 20 -> 10 -> null

B. 40 -> 30 -> 20 -> 25 -> 10 -> null

C. 40 -> 30 -> 20 -> 10 -> 25 -> null

D. 40 -> 25 -> 30 -> 20 -> 10 -> null

Solution

1. Step 1: Reverse the original list

   Original: 10 -> 20 -> 30 -> 40 -> null reversed becomes 40 -> 30 -> 20 -> 10 -> null.

2. Step 2: Insert 25 between 30 and 20

   In reversed list, 30 points to 20, so insert 25 after 30: 30 -> 25 -> 20.

3. Step 3: Final list order

   40 -> 30 -> 25 -> 20 -> 10 -> null.

4. Final Answer:

   40 -> 30 -> 25 -> 20 -> 10 -> null -> Option A

5. Quick Check:

   Insert between 30 and 20 in reversed list = 40->30->25->20->10 [OK]

Hint: Reverse first, then insert at correct reversed position [OK]

Common Mistakes:

• Inserting 25 between wrong nodes after reversal
• Confusing original and reversed order
• Not updating next pointers correctly