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

Time Complexity: Reverse a Singly Linked List Iterative

O(n)

Understanding Time Complexity

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

How many steps does it take to reverse all the links in the list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function reverseLinkedList(head) {
  let prev = null;
  let current = head;
  while (current !== null) {
    let next = current.next;
    current.next = prev;
    prev = current;
    current = next;
  }
  return prev;
}
    

This code reverses the links of a singly linked list by changing each node's next pointer.

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

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: 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 linearly with the number of nodes.

Common Mistake

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

[OK] Correct: Each node's pointer must be changed one by one, so the time depends on the list size.

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 does reversing a singly linked list iteratively mean?
head -> 1 -> 2 -> 3 -> null becomes?

easy

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

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

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

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

Solution

1. Step 1: Understand the linked list order

   The original list goes from 1 to 3 in order: 1 -> 2 -> 3 -> null.

2. Step 2: Reverse the pointers iteratively

   Reversing means the last node becomes first, so the list becomes 3 -> 2 -> 1 -> null.

3. Final Answer:

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

4. Quick Check:

   Reverse order = head -> 3 -> 2 -> 1 -> null [OK]

Hint: Reversing flips the list order completely [OK]

Common Mistakes:

• Thinking the list stays the same after reverse
• Confusing node values with pointers
• Assuming null moves to front

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

easy

A. let prev = null, current = null;

B. let prev = head, current = null;

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

D. let prev = null, current = head;

Solution

1. Step 1: Initialize previous pointer to null

   Before reversal, previous node is null because head has no previous.

2. Step 2: Initialize current pointer to head

   Current starts at head to traverse the list.

3. Final Answer:

   let prev = null, current = head; -> Option D

4. Quick Check:

   Start prev=null, current=head [OK]

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

Common Mistakes:

• Setting prev to head initially
• Setting current to null at start
• Swapping initialization order

3. What is the output of the following code?

class Node {
  constructor(val) {
    this.val = val;
    this.next = null;
  }
}

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

// Create list: 1 -> 2 -> 3 -> null
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

let reversedHead = reverseList(head);

let result = [];
let node = reversedHead;
while (node) {
  result.push(node.val);
  node = node.next;
}
console.log(result);

medium

A. []

B. [3, 2, 1]

C. [null]

D. [1, 2, 3]

Solution

1. Step 1: Trace reversal loop

   Each iteration reverses the next pointer: 1->2->3 becomes 3->2->1.

2. Step 2: Collect values from reversed list

   After reversal, traversing from prev (new head) gives values 3, 2, 1.

3. Final Answer:

   [3, 2, 1] -> Option B

4. Quick Check:

   Reversed list values = [3, 2, 1] [OK]

Hint: Reversed list values appear in reverse order [OK]

Common Mistakes:

• Returning original head instead of prev
• Not updating current pointer correctly
• Collecting values before reversal

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

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

medium

A. head.next should not be set to null

B. current should start as head, not head.next

C. prev should start as null, not head

D. nextTemp should be assigned after changing current.next

Solution

1. Step 1: Check initial pointer setup

   prev must start as null because head has no previous node before reversal.

2. Step 2: Understand impact of wrong prev initialization

   Starting prev as head causes a cycle or incorrect links during reversal.

3. Final Answer:

   prev should start as null, not head -> Option C

4. Quick Check:

   prev=null start fixes reversal [OK]

Hint: Prev must start null to avoid cycles [OK]

Common Mistakes:

• Starting prev at head causing loops
• Not setting head.next to null after reversal
• Changing nextTemp after modifying current.next

5. Given a singly linked list with nodes containing values 1 -> 2 -> 3 -> 4 -> 5 -> null, write an iterative function to reverse only the first 3 nodes. What will be the list after reversal?

hard

A. 3 -> 2 -> 1 -> 4 -> 5 -> null

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

C. 5 -> 4 -> 3 -> 2 -> 1 -> null

D. 1 -> 2 -> 3 -> 4 -> 5 -> null

Solution

1. Step 1: Reverse first 3 nodes iteratively

   Reverse nodes 1, 2, 3 to get 3 -> 2 -> 1.

2. Step 2: Connect reversed part to remaining list

   Node 1 (now last in reversed part) points to node 4, preserving rest of list.

3. Final Answer:

   3 -> 2 -> 1 -> 4 -> 5 -> null -> Option A

4. Quick Check:

   Partial reversal keeps rest intact [OK]

Hint: Reverse only first k nodes, link rest after [OK]

Common Mistakes:

• Reversing entire list instead of first 3 nodes
• Not reconnecting reversed part to rest
• Breaking list by losing tail connections