3Sum
Given an integer array `nums`, return all the triplets `[nums[i], nums[j], nums[k]]` such that $i \neq j$, $i \neq k$, and $j \neq k$, and `nums[i] + nums[j] + nums[k] == 0`.
Why Interviewers Ask This
Airbnb asks the 3Sum problem to evaluate a candidate's ability to optimize brute-force solutions using sorting and the two-pointer technique. They specifically look for proficiency in handling edge cases like duplicate triplets and managing time complexity constraints efficiently, which mirrors their need for scalable, high-performance code in real-world booking systems.
How to Answer This Question
1. Clarify requirements immediately by confirming if duplicates are allowed in output and what happens with negative numbers. 2. Propose a naive O(n^3) solution first to establish a baseline, then immediately pivot to an optimized approach. 3. Explain your strategy: sort the array first to enable the two-pointer technique, reducing complexity to O(n^2). 4. Walk through the logic of fixing one number and searching for two others that sum to its negation using left and right pointers. 5. Crucially, detail your deduplication logic to skip identical values during iteration, ensuring no duplicate triplets appear in the result set, as Airbnb values clean, bug-free data processing.
Key Points to Cover
- Demonstrating the transition from O(n^3) brute force to O(n^2) optimization via sorting
- Explicitly detailing the logic used to skip duplicate values and prevent redundant triplets
- Explaining the two-pointer movement strategy based on whether the current sum is too low or too high
- Acknowledging edge cases such as empty arrays or arrays with fewer than three elements
- Maintaining clear communication about time and space complexity throughout the explanation
Sample Answer
To solve the 3Sum problem efficiently, I would start by sorting the input array, which takes O(n log n) time. This allows us to use the two-pointer technique effectively. We iterate through the array with a fixed pointer 'i', treating nums[i] as the first element of our triplet. For each 'i', we initialize two pointers: 'left' at i + 1 and 'right' at the end of the array. We calculate the sum of nums[i], nums[left], and nums[right]. If the sum equals zero, we add the triplet to our results. To handle duplicates, which is critical for correctness, we must ensure that if the current 'i' is the same as the previous one, we skip it entirely. Similarly, after finding a valid pair, we move 'left' forward and 'right' backward while skipping any adjacent duplicate values to avoid adding the same triplet multiple times. If the sum is less than zero, we increment 'left' to increase the total; if greater, we decrement 'right'. This approach ensures we find all unique triplets in O(n^2) time complexity, which is optimal for this constraint set. This method demonstrates robustness against edge cases like arrays with many zeros or repeated elements, ensuring the output remains clean and accurate.
Common Mistakes to Avoid
- Failing to sort the array first, which makes the two-pointer technique impossible to implement correctly
- Missing the deduplication logic, leading to duplicate triplets in the final output list
- Not handling the case where the array has fewer than three elements before starting the loop
- Using nested loops without optimizing, resulting in Time Limit Exceeded errors on large inputs
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