Leetcode #1836: Remove Duplicates From an Unsorted Linked List
In this guide, we solve Leetcode #1836 Remove Duplicates From an Unsorted Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given the head of a linked list, find all the values that appear more than once in the list and delete the nodes that have any of those values. Return the linked list after the deletions.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Hash Table, Linked List
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: head = [1,2,3,2]
Output: [1,3]
Explanation: 2 appears twice in the linked list, so all 2's should be deleted. After deleting all 2's, we are left with [1,3].
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def deleteDuplicatesUnsorted(self, head: ListNode) -> ListNode:
cnt = Counter()
cur = head
while cur:
cnt[cur.val] += 1
cur = cur.next
dummy = ListNode(0, head)
pre, cur = dummy, head
while cur:
if cnt[cur.val] > 1:
pre.next = cur.next
else:
pre = cur
cur = cur.next
return dummy.next
Complexity
The time complexity is , and the space complexity is , where is the length of the linked list. The space complexity is , where is the length of the linked list.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.