Remove Duplicates from Sorted List

To solve this problem efficiently, I would first verify the base cases. If the head is null or contains only one node, we simply return the head as no duplicates can exist. Since the list is already sorted, any duplicates will be adjacent to each other, which allows us to solve this in a single linear pass with constant space complexity, aligning with Amazon's preference for optimized solutions. I will use a single pointer approach. Starting at the head, I'll iterate through the list comparing the current node's value with the next node's value. If they are identical, it means we have found a duplicate. Instead of creating a new node or using a hash set, I will simply bypass the duplicate by pointing the current node's next reference to the node after the duplicate. This effectively removes the duplicate from the chain. If the values differ, I simply advance my pointer to the next node to continue the check. For example, given the list 1 -> 1 -> 2, the pointer starts at the first 1. It sees the next node is also 1. I update the first node's next pointer to skip the second 1 and point directly to 2. The list becomes 1 -> 2. Moving forward, 1 and 2 are different, so I advance. The loop ends, and I return the modified head. This ensures the result remains sorted and contains unique elements without extra memory overhead.