The naive sort-based solution is O(n log n) and I kept second-guessing whether they'd accept it.
Use a hash set to store all elements for O(1) lookups. For each number, check if it's the start of a sequence (i.e., num-1 not in set), then count consecutive numbers. This ensures each element is visited at most twice, achieving O(n) time.
Pro tip: Clarify that the O(n) time complexity relies on hash set operations being O(1) on average, and mention that the space complexity is O(n) due to the set. Also, discuss edge cases like empty array or duplicates.
Restate the problem: find the length of the longest consecutive sequence in an unsorted array, with O(n) time. Confirm that elements can be in any order and duplicates may exist.
Use a hash set to store all unique elements, enabling O(1) average-time membership checks. This is key to achieving linear time.
For each number in the set, check if it is the start of a sequence by verifying that num-1 is not in the set. Only start counting from these numbers to avoid redundant work.
From each start, incrementally check for num+1, num+2, etc., in the set, counting the length. Update the maximum length found.
Explain that each element is visited at most twice (once when checking if it's a start, once when counting), so time is O(n). Space is O(n) for the set.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.