8

The problem requires us to iterate a sorted list and remove any duplicates possible. It’s a straightforward problem, but don’t get confused like I did. I did not read the “sorted” and thought it was a regular list. So I used a set to keep track of duplicates.

BUT, since it’s a sorted list, one can just iterate once, and see if the next value is the same as the current one, if yes → skip the nodes.

Code:

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        # LIST IS SORTED BRO, SO JUST CHECK CONSEQUTIVELY
 
        n = head
        while n:
            while n.next and n.next.val == n.val:
                n.next = n.next.next
 
            n = n.next
 
        return head

Big O Analysis

• Runtime The runtime complexity here is O(N) , regardless of whether duplicates are present or not we would still have to visit each node atleast once.
• Memory The memory usage is O(1) since we do not use any extra data structure.

— A

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