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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Bytedance SWE interview with a classic algorithmic problem. Nothing too wild but the O(n^2) constraint made it clear they wanted a specific approach, not just any working solution.

Questions Asked (1)

Q1

Given an integer array, find all unique triplets that sum to zero. No duplicate triplets in the result, and you need to do it in O(n^2) time.

Algorithms & Data Structures
Author's notes

My first instinct was a brute force triple loop which obviously wasn't going to fly.

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AI HintsAI Generated

Suggested Approach

Start by sorting the array to enable efficient two-pointer traversal and easy duplicate skipping. Then, iterate through each element as the first number of the triplet, and for each, use two pointers to find pairs that sum to the negative of that number. Skip duplicates at all levels to ensure unique triplets.

Pro tip: Explicitly discuss time and space complexity: O(n^2) time due to nested loops (sorting O(n log n) is dominated), and O(1) extra space if output not counted. Also, mention that sorting modifies input; if not allowed, copy first.

1. Clarify and Sort

Confirm input constraints (e.g., array size, possible duplicates) and sort the array in ascending order. Sorting is key for two-pointer technique and duplicate handling.

2. Iterate with Two Pointers

Loop through the array with index i from 0 to n-3. For each i, set left = i+1 and right = n-1, and compute sum = nums[i] + nums[left] + nums[right].

3. Adjust Pointers

If sum < 0, increment left; if sum > 0, decrement right; if sum == 0, record triplet and move both pointers while skipping duplicates.

4. Skip Duplicates

After finding a triplet, skip duplicate values for left and right. Also, in the outer loop, skip duplicate values for i to avoid duplicate triplets.

5. Return Result

Collect all unique triplets in a list and return it. Ensure no duplicate triplets are included.

Key Points to Mention

  • Sorting the array first to enable two-pointer technique and duplicate skipping.
  • Using two pointers (left and right) to find pairs that sum to the negative of the current element.
  • Skipping duplicates at all levels (outer loop and inner while loops) to ensure unique triplets.
  • Time complexity: O(n^2) due to nested loops (sorting is O(n log n), which is dominated).
  • Space complexity: O(1) extra space if output is not counted, or O(n) for sorting if in-place not allowed.
  • Handling edge cases: empty array, array with fewer than 3 elements, all zeros, etc.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.