Longest Consecutive Sequence
Given an unsorted array of integers `nums`, return the length of the longest consecutive elements sequence. The algorithm should run in $O(n)$ time.
Why Interviewers Ask This
Tesla evaluates candidates on algorithmic efficiency and the ability to handle large-scale data processing within strict constraints. This question specifically tests if you can distinguish between sorting (O(n log n)) and hashing (O(n)), proving you understand time complexity trade-offs essential for real-time vehicle systems where performance is critical.
How to Answer This Question
1. Clarify requirements immediately, confirming the O(n) constraint and handling edge cases like empty arrays or duplicates. 2. Propose a brute-force solution first to show baseline understanding, then explain why it fails the time limit. 3. Pivot to the optimal approach using a Hash Set for O(1) lookups; explain that you only start counting sequences from numbers where 'num-1' does not exist to avoid redundant work. 4. Walk through the logic step-by-step with a dry run example, such as [100, 4, 200, 1, 3, 2], demonstrating how the set skips non-starting elements. 5. Analyze the final complexity, explicitly stating why the nested loop structure still results in linear time because each element is visited at most twice.
Key Points to Cover
- Explicitly acknowledging the O(n) constraint prevents wasting time on sorting solutions
- Explaining the optimization of skipping non-sequence-starts demonstrates deep algorithmic insight
- Using a Hash Set for O(1) lookup is the critical technical enabler for this solution
- Providing a concrete dry run with specific numbers validates the logic clearly
- Correctly analyzing why the nested loops do not violate the linear time complexity
Sample Answer
To solve the Longest Consecutive Sequence problem efficiently within Tesla's high-performance standards, I would prioritize an O(n) time complexity solution. First, I would convert the input array into a Hash Set to allow for constant-time lookups. The core insight here is that we don't need to check every number as a potential start of a sequence. We only initiate a search if the current number is the beginning of a sequence, meaning the number minus one is not present in our set.
For example, given the array [100, 4, 200, 1, 3, 2], I would insert all elements into the set. When iterating through 100, since 99 isn't there, I know it starts a sequence. I would then check 101, find it missing, and record a length of 1. Next, when I encounter 4, 3 is in the set, so I skip it. However, when I hit 1, since 0 is absent, I start counting: 1 exists, 2 exists, 3 exists, but 4 is already checked. Wait, 4 is in the set, so the sequence is 1, 2, 3, 4. I continue until 5 is missing, recording a length of 4. By ensuring we only build sequences from their true starting points, we guarantee that every element is accessed at most twice—once during iteration and once during the sequence expansion. This approach ensures we meet the strict O(n) requirement while maintaining clean, readable code suitable for production environments.
Common Mistakes to Avoid
- Attempting to sort the array first, which results in O(n log n) time and violates the prompt
- Checking every number to start a sequence, leading to O(n^2) worst-case complexity
- Failing to handle duplicate numbers correctly by not converting the input to a Set first
- Not explaining why the algorithm remains linear despite having two nested loops in the code
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