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Bytedance·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate
May 2026

Summary

Bytedance coding round for a Software Engineer role. Two algorithm problems, both leaning toward the harder side of what you'd expect for this kind of screen.

Questions Asked (2)

Q1

Given a list of closed intervals, merge all overlapping ones and return a sorted list of non-overlapping intervals.

Algorithms & Data Structures
Author's notes

Classic problem but I fumbled the edge case where one interval completely swallows the next.

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AI HintsAI Generated

Suggested Approach

Start by sorting the intervals based on their start times, then iterate through them while maintaining a result list of merged intervals. For each interval, if it overlaps with the last merged interval, merge them by updating the end time; otherwise, add it to the result.

Pro tip: Clarify edge cases upfront, such as empty input, single interval, or intervals with the same start but different ends. Also, mention that the solution runs in O(n log n) time due to sorting, which is optimal for comparison-based approaches.

1. Understand the problem and edge cases

Confirm that intervals are closed and may be unsorted. Discuss edge cases like empty list, single interval, and intervals that touch (e.g., [1,2] and [2,3] are considered overlapping).

2. Sort intervals by start time

Sort the input list of intervals based on their start values. This ensures that any overlapping intervals will be adjacent, simplifying the merging process.

3. Iterate and merge

Initialize an empty result list. For each interval in the sorted list, if the result list is empty or the current interval does not overlap with the last interval in the result, append it. Otherwise, merge by updating the end of the last interval to the maximum of its end and the current interval's end.

4. Return the merged list

After processing all intervals, return the result list, which contains non-overlapping intervals sorted by start time.

5. Analyze complexity and test

State that time complexity is O(n log n) due to sorting, and space complexity is O(n) for the output. Walk through a few test cases to verify correctness.

Key Points to Mention

  • Sorting is crucial to bring overlapping intervals together, enabling a linear scan.
  • Overlap condition: two intervals overlap if the start of the current interval is less than or equal to the end of the previous merged interval.
  • When merging, the new end is the maximum of the two ends to handle cases where one interval is completely contained within another.
  • Time complexity: O(n log n) due to sorting; space complexity: O(n) for the output list (or O(1) extra if modifying in place, but typically O(n)).
  • Edge cases: empty input, single interval, intervals that are already non-overlapping, and intervals that touch at endpoints.
  • The algorithm is optimal for comparison-based sorting, and no better than O(n log n) is possible in the general case.

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

Q2

Given an array of bar heights in a histogram, find the maximum rectangular area you can form using one or more consecutive bars.

Algorithms & Data Structures
Author's notes

This one hurt.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then propose a brute-force O(n^2) solution as a baseline. Follow up with an optimized O(n) approach using a monotonic stack to find the largest rectangle in linear time, explaining the intuition and handling edge cases.

Pro tip: Emphasize that the stack stores indices of bars in increasing height order, and that you can avoid sentinel values by handling the final flush carefully—this shows attention to code robustness and edge cases.

1. Clarify and Restate

Confirm the problem: given an array of bar heights, find the maximum rectangular area formed by consecutive bars. Discuss edge cases like empty array, single bar, and all equal heights.

2. Brute Force Baseline

Describe the O(n^2) approach: for each bar, expand left and right until a shorter bar is found, computing area. This sets a baseline and shows you can think simply first.

3. Optimized Stack Approach

Introduce the monotonic stack: iterate through bars, maintain a stack of indices with increasing heights. When a shorter bar is encountered, pop and calculate area with the popped bar as the smallest height.

4. Handle Final Flush

After iteration, pop remaining bars in the stack, treating the right boundary as the end of the array. Compute areas similarly.

5. Complexity and Edge Cases

State time and space complexity: O(n) time, O(n) space. Discuss edge cases like empty input, single bar, and decreasing/increasing heights.

Key Points to Mention

  • Monotonic stack maintains indices of bars in increasing height order.
  • Area calculation: height of popped bar * (current index - stack top index - 1).
  • Time complexity O(n) because each bar is pushed and popped at most once.
  • Space complexity O(n) for the stack.
  • Edge cases: empty array, single bar, all bars same height, strictly increasing/decreasing heights.
  • Alternative approaches: divide and conquer, but stack is optimal.

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