Free AI LeetCode Communication Coach — Reverse Linked List

The problem

Given the head of a singly linked list, reverse the list and return the new head.

Example: 1 → 2 → 3 → 4 → 5 → 5 → 4 → 3 → 2 → 1

Say you save next before rewiring.

prev = None
curr = head
while curr:
    nxt = curr.next
    curr.next = prev
    prev = curr
    curr = nxt
return prev

How to explain it

1. 1. Restate

   Say the problem in your own words.

   One or two sentences. Show you understood the input, the output, and the goal — not that you memorised the prompt.

2. 2. Approach

   Name the method before you code.

   Brute force first if you need it, then the structure you will use: hash map, two pointers, stack, binary search.

3. 3. Example

   Walk one concrete input.

   Pick small numbers. Say what you store, what you compare, and what you return. Interviewers follow an example more easily than abstract talk.

4. 4. Time and space

   One sentence each.

   After the example, before you claim you are done. “Time is O(n) because we scan once. Space is O(n) for the map.”

5. 5. Edge cases

   Name at least one unusual input.

   Empty input, duplicates, already sorted, overflow. Invite a follow-up: “I would also check …”

• “I keep prev, curr, and next.”
• “I reverse the link before I move forward.”

Practise out loud

1. 1. Restate

   I need to reverse every next pointer and return the new head.

2. 2. Approach

   I will iterate with three pointers: previous, current, and next. I reverse the link, then advance.

3. 3. Example

   On 1→2→3, I reverse 1 to null, then 2 to 1, then 3 to 2. The new head is 3.

4. 4. Time and space

   Time is O(n). Space is O(1).

5. 5. Edge cases

   I would check null head and a single-node list.

Common mistakes when explaining Reverse Linked List

• Losing the next pointer before rewiring.
• Forgetting to return the new head (old tail).