Coding Interview Preparation on Telegram: “🎯 CODING CHALLENGE #3

🎯 CODING CHALLENGE #3 - Reverse a Linked List
Difficulty: Easy-Medium | Asked at: Amazon, Apple, Adobe

Reverse a singly linked list, iteratively.


Input:  1 → 2 → 3 → 4 → None
Output: 4 → 3 → 2 → 1 → None

💡 Hint: You need to track three pointers as you walk the list: the previous node, the current node, and the next node - because once you flip a pointer, you lose the way forward unless you saved it first.

Solution:

python
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverse_list(head):
    prev = None
    curr = head
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    return prev

Complexity: O(n) time, O(1) space - this is the detail that separates a strong answer from an average one. A recursive solution is O(n) time but O(n) space due to the call stack - know both, and be ready to explain the tradeoff.

Common mistake: Forgetting to save curr.next before overwriting it, which permanently disconnects the rest of the list.

Iterative or recursive - which do you reach for first, and why? 👇