This is a premium problem. We're working on making it available for free soon.
Use these hints if you're stuck. Try solving on your own first.
Is there a way we can know beforehand which nodes to delete?
Count the number of appearances for each number.
Solutions for this premium problem will be available for free soon.
Browse Free ProblemsWatch expert explanations and walkthroughs
Practice problems asked by these companies to ace your technical interviews.
Explore More ProblemsJot down your thoughts, approach, and key learnings
Yes, variations of this problem frequently appear in technical interviews at large tech companies. Interviewers use it to evaluate understanding of linked lists, hashing, pointer manipulation, and trade-offs between time and space complexity.
A hash set is the most effective data structure for this problem. It allows quick lookups to check whether a value has already appeared in the list, enabling efficient duplicate removal during a single pass.
The optimal approach uses a hash set to track values already seen during traversal. As you iterate through the linked list, remove any node whose value already exists in the set. This allows duplicate detection in constant time and results in O(n) overall time complexity.
Yes, it can be solved without additional memory by using a nested traversal approach. For each node, you scan the remaining list and remove nodes with the same value. However, this increases the time complexity to O(n²).